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MATHS 5 Marks Questions

Linear programming · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 116 questions.

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Question 15 Marks
A wholesale dealer deals in two kinds, A and B (say) of mixture of nuts. Each kg of mixture A contains 60 grams of almonds, 30 grams of cashew nuts and 30 grams of hazel nuts. Each kg of mixture B contains 30 grams of almonds, 60 grams of cashew nuts and 180 grams of hazel nuts. The remainder of both mixtures is per nuts. The dealer is contemplating to use mixtures A and B to make a bag which will contain at least 240 grams of almonds, 300 grams of cashew nuts and 540 grams of hazel nuts. Mixture A costs Rs. 8 per kg. and mixture B costs Rs. 12 per kg. Assuming that mixtures A and B are uniform, use graphical method to determine the number of kg. of each mixture which he should use to minimise the cost of the bag.
Answer
Let required number of bag A and bag B be x and y respectively.
Since, costs of each bag A and bag B are Rs. 8 and Rs. 12 per kg.
So, cost of x number of bag A and y number of bag B are Rs. 8x and Rs. 12y respectively,Let z be total cost of bags, so,
$Z = 8x + 12y$
Since, each bag A and B contain 60 and 30gms of almonds respectively.
So, x bags of A and y bags of B contain 60x and 30ygms. of almonds respectively but, mixtures should contain at least 240gms almonds, so,
$60\text{x}+30\text{y}\geq240$
$2\text{x}+\text{y}\geq8$ (first constraint)
Since, each bag A and B contain 30 and 60 gms of cashew nuts respectively.
So, x bags of A and y bags of B contain 30x and 60y gms. of cashew nuts respectively but, mixtures should contain at least 300 gms of cashew nuts, so
$30\text{x}+60\text{y}\geq300$
$\text{x}+2\text{y}\geq10$ (second constraint)
Since, each bag A and B contain 30 and 180 gms. of hazel nuts respectively.
So, x bags of A and y bags of B contain 30x and 180y gms. of hazel nuts respectively but, mixtures should contain at least 540gms of hazel nuts, so,
$30\text{x}+180\text{y}\geq540$
$\text{x}+6\text{y}\geq18$ (third constraint)
Hence, mathematical formulation of LPP is,
Find x and y which maximize
$Z = 8x + 12y$
Subject to constraints,
$2\text{x}+\text{y}\geq8$
$\text{x}+2\text{y}\geq10$
$\text{x}+6\text{y}\geq18$
$\text{x},\text{y}\geq0$ [Since quantity of bags can not be less than zero]
Region $2\text{x}+\text{y}\geq8$:
line $2x + y = 8 $ meets axes at $A_1(4, 0), B_1(0, 8)$ respectively.
Region not containing origin represents $2\text{x}+\text{y}\geq8$ as $(0, 0)$ does not satisfy $2\text{x}+\text{y}\geq8$.
Region $\text{x}+2\text{y}\geq10$:
line $x + 2y = 10$ meets axes at $A_2(10, 0), B_2(0, 5)$ respectively.
Region not containing origin represents $\text{x}+2\text{y}\geq10$ as $(0, 0)$ does not satisfy $\text{x}+2\text{y}\geq10$.
Region $\text{x}+6\text{y}\geq18$:
line $x + 6y = 18$ meets axes at $A_3(18, 0), B_3(0, 3)$ respectively.
Region not containing origin represents $\text{x}+6\text{y}\geq18$ as $(0, 0)$ does not satisfy $\text{x}+6\text{y}\geq18$.
Region $\text{x},\text{y}\geq0$: it represents first quadrant.

Unbouded shaded region $A_3PQB_1$ is feasible region with corner point $A_3(18, 0), P (6, 2) Q(2, 4), B_1(0, 8).$
P is obtained by solving $x + 6y = 18$ and $x + 2y = 10, Q$ is obtained by solving $2x + y = 8$ and $x + 2y = 10$
The value of $z = 8x +12y$ at
$A_3(18, 0) = 8(18) +12(0) = 144$
$P(6, 2) = 8(6) + 12(2) = 72$
$Q(2, 4) = 8(2) + 12(4) = 64$
$B_1(0,8) = 8(0) + 12(8) = 96$
Smallest value of Z is 64, open half plane $​​8\text{x}+12\text{y}\geq64$ has no point is common with feasible region, so, smallest value is the minimum value
Minimum cost = Rs. 64
quantity of mixture A = 2kg.
quantity of mixture 8 = 4kg.
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Question 25 Marks
A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs. 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs. 200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
Answer
Let x bags of brand P and y bags of brand Q should be mixed to produce the mixture.

Each bag of brand P costs Rs. 250 and each bag of brand Q costs Rs. 200.

Therefore, x bags of brand P and y bags of brand Q costs Rs. (250x + 200y).

Since each bag of brand P contains 3 units of nutritional element A and each bag of brand Q contains 1.5 units of nutritional element A, therefore, x bag of brand P and y bag of brand Q will contain (3x + 1.5y) units of nutritional element A.

But, the minimum requirement of nutrients A is 18 units.

$3\text{x}+1.5\text{y}\geq18$

$2\text{x}+\text{y}\geq12$

Similarly, x bag of brand P and y bag of brand Q will contain (2.5x + 11.25y) units of nutritional element B.

But, the minimum requirement of nutrients B is 45 units.

$2.5\text{x}+11.25\text{y}\geq45$

$2\text{x}+9\text{y}\geq36$

Also, x bag of brand P and y bag of brand Q will contain (2x + 3y) units of nutritional element B.

But, the minimum requirement of nutrients C is 24 units.

$2\text{x}+3\text{y}\geq24$

Thus, the given linear programming problem is,

Minimise Z = 250x + 200y

Subject to the constraints

$2\text{x}+\text{y}\geq12$

$2\text{x}+9\text{y}\geq36$

$2\text{x}+3\text{y}\geq24$

$\text{x},\text{y}\geq0$

The feasible region determined by the given constraints can be diagrammatically represented as,



The coordinates of the corner points of the feasible region are A(18, 0), B(9, 2), C(3, 6) and D(0, 12).

The value of the objective function at these points are given in the following table.

= 250x+ 200y
Corner Point
Z = 250x + 200y
(18, 0)
50 × 18 + 200 × 0 = 4500
(9, 2)
250 × 9 + 200 × 2 = 2650
(3, 6) 250 × 3 + 200 × 6 = 1950 → Minimum
(0, 12) 250 × 0 + 200 × 12 = 2400
The smallest value of Z is 1950 which is obtained at (3, 6).

It can be seen that the open half-plane represented by 250x + 200y < 1950 or 5x + 4y < 39 has no common points with the feasible region.

So, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimise the cost.

Hence, the minimum cost of the mixture per bag is 1950.
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Question 35 Marks
An aeroplane can carry a maximum of 200 passengers. A profit of Rs.1000 is made on each executive class ticket and a profit of Rs.600 is made on each economy class ticket. The airline reserves atleast 20 seats for executive class. However, atleast 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit of the airline. What is the maximum profit?
Answer
Suppose x tickets of executive class and y tickets of economy class are sold by the airline.
The profit on each executive class ticket is Rs. 1000 and on each economy class ticket is Rs.600.

Therefore, the total profit from x executive class tickets and y economy class ticket is Rs.(1000x + 600y).

Now, the aeroplane can carry a maximum of 200 passengers.

x + y ≤ 200

The airline reserves atleast 20 seats for executive class.

x ≥ 20

Also, atleast 4 times as many passengers prefer to travel by economy class than by the executive class.

y ≥ 4x

Thus, the given linear programming problem is

Maximise Z = 1000x + 600y

Subject to the constraints

x + y ≤ 200

x ≥ 20

y ≥ 4x

x, y ≥ 0

The feasible region determined by the given constraints can be diagrammatically represented as,



The coordinates of the corner points of the feasible region are A(20, 80), B(40, 160) and C(20, 180).

The value of the objective function at these points are given in the following table.
Corner Point
Z =1000x + 600y
(20, 8)
1000 × 20 + 600 × 80 = 68000
(40, 10)
1000 × 40 + 600 × 160 = 136000 → Maximum
(20, 180)
1000 × 20 + 600 × 180 =128000
The maximum value of Z is 136000 at x = 40, y = 160.

Hence, 40 tickets of executive class and 160 tickets of economy class should be sold to maximise the profit.

The maximum profit of the airline is Rs. 1,36,000.
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Question 45 Marks
A dietician mixes together two kinds of food in such a way that the mixture contains at least 6 units of vitamin A, 7 units of vitamin B, 11 units of vitamin C and 9 units of vitamin D. The vitamin contents of 1kg of food X and 1kg of food Y are given below:
 
Vitamin
A
Vitamin
B
Vitamin
C
Vitamin
D
Food X
1
1
1
2
Food Y
2
1
3
1
One kg food X costs Rs. 5, whereas one kg of food Y costs Rs. 8.
Find the least cost of the mixture which will produce the desired diet.
Answer
Let the dietician wishes to mix x kg of food X and y kg of Y.
Therefore,
As we are given,
 
Vitamin
A
Vitamin
B
Vitamin
C
Vitamin
D
Food X
1
1
1
2
Food Y
2
1
3
1
It is given that the mixture should contain at least 6 units of vitamin A, 7 units of vitamin B, 11 units of vitamin C and 9 units of vitamin D.
Therefore, the constrants are
$\text{x}+2\text{y}\geq6$
$\text{x}+\text{y}\geq7$
$\text{x}+3\text{y}\geq11$
$2\text{x}+\text{y}\geq9$
It is given that cost of food X is Rs 5 per kg and cost of food Y is Rs. 8 per kg.
Thus, $Z= 5x+8y$
Thus, the mathematical formulation of the given linear programmimg problem is
Minimize $Z= 5x+8y$
subject to
a
$\text{x}+2\text{y}\geq6$
$\text{x}+\text{y}\geq7$
$\text{x}+3\text{y}\geq11$
$2\text{x}+\text{y}\geq9$
First, we will convert the given inequations into equations, we obtain the following equations:
$x + 2y = 6, x + y = 7, x + 3y = 11, 2x + y =9, x = 0$ and $y=0.$
The line x + 2y = 6 meets the coordinate axis at $A_1(6, 0)$ and $B_1(0, 3)$.
Join these points to obtain the line $x + 2y = 6.$
Clearly, (0, 0) does not satisfies the inequation $\text{x}+2\text{y}\geq6$.
So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
The line $x + y = 7$ meets the coordinate axis at $C_1(7, 0)$ and $D_1(0, 7).$
Join these points to obtain the line $x + y = 7.$
Clearly, $(0, 0)$ does not satisfies the inequation $\text{x}+\text{y}\geq7$.
So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
The line $x + 3y = 11$ meets the coordinate axis at $E_1(11, 0)$ and $\text{F}_1\Big(0,\frac{11}{3}\Big)$.
Join these points to obtain the line $x + 3y = 11.$
Clearly, $(0, 0)$ does not satisfies the inequation $\text{x}+3\text{y}\geq11$.
So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
The line $2x + y = 9$ meets the coordinate axis at $\text{G}_1\Big(\frac{9}{2},0\Big)$ and $H(0, 9).$
Join these points to obtain the line $2x + y = 9.$
Clearly, $(0, 0)$ does not satisfies the inequation $2\text{x}+\text{y}\geq9$.
So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
Region represented by $\text{x}\geq0$ and $\text{y}\geq0$:
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The feasible region determined by the system of constraints is

The corner points are $H_1(0, 9), I_1(2, 5), J_1(5, 2), E_1(11, 0).$
The values of Z at these corner points are as follows
Corner point
Z =5x + 8y
$H_1$
72
$I_1$
50
$J_1$
41
$E_1$
55
The minimum value of Z is at $J_1(5, 2)$ which is Rs. 41.
Hence, cheapest combination of foods will be 5 units of food X and 2 units of food Y.
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Question 55 Marks
Maximum Z = 5x + 3y Subject to $2\text{x}+\text{y}\geq10$ $\text{x}+3\text{y}\geq15$ $\text{x}\leq10$$\text{y}\leq8$
$\text{x},\text{y}\geq0$
Answer
First, we will convert the given inequations into equations, we obtain the following equations:

2x + y = 10, x + 3y = 15, x = 10, y = 8

Region represented by $2\text{x}+\text{y}\geq10$:

The line 2x + y = 10 meets the coordinate axes at A(5, 0) and B(0, 10) respectively.

By joining these points we obtain the line 2x + y = 10. Clearly (0, 0) does not satisfies the inequation $2\text{x}+\text{y}\geq10$.

So,the region in xy plane which does not contain the origin represents the solution set of the inequation $2\text{x}+\text{y}\geq10$.

Region represented by $\text{x}+3\text{y}\geq15$:

The line x + 3y = 15 meets the coordinate axes at C(15, 0) and D(0, 5) respectively.

By joining these points we obtain the line x + 3y = 15.

Clearly (0, 0) satisfies the inequation $\text{x}+3\text{y}\geq15$. o, the region in xy plane which does not contain the origin represents the solution set of the inequation $\text{x}+3\text{y}\geq15$.

The line x = 10 is the line that passes through the point (10, 0) and is parallel to Y axis. $\text{x}\leq10$ is the region to the left of the line x = 10.

The line y = 8 is the line that passes through the point (0, 8) and is parallel to X axis. $\text{y}\leq8$ is the region below the line y = 8.

Region represented by $\text{x}\geq0$ and $\text{y}\geq0$:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations $\text{x}\geq0$ and $\text{y}\geq0$.

The feasible region determined by the system of constraints, $2\text{x}+\text{y}\geq10$, $\text{x}+3\text{y}\geq15$, $\text{x}\leq10$, $\text{y}\leq8$, $\text{x}\geq0$ and $\text{y}\geq0$ are as follows.



The corner point of the feasible region are E(3, 4), $\text{H}\Big(10,\frac{5}{3}\Big),$ F(10, 8) and G(1, 8).

The values of Z at these corner points are as follows.
$\text{Corner point}$
$\text{Z}=5\text{x}+3\text{y}$
$\text{E}(3, 4)$
$5\times3+3\times4=27$
$\text{H}\Big(10,\frac{5}{3}\Big)$
$5\times10+3\times\frac{5}{3}=55$
$\text{F}(10, 8)$
$5\times10+3\times8=74$
$\text{G}(1, 8)$
$5\times1+3\times8=29$
Therefore, the minimum value of Z is 27 at the point F(3, 4).

Hence, x = 3 and y = 4 is the optimal solution of the given LPP.

Thus, the optimal value of Z is 27.
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Question 65 Marks
A firm manufactures two types of products $A$ and $B$ and sells them at a profit of Rs 2 on type $A$ and Rs 3 on type $B$. Each product is processed on two machines $M_1$ and $M_2$. Type $A$ requires one minute of processing time on $M_1$ and two minutes of $M_2$; type $B$ requires one minute on $M_1$ and one minute on $M_2$. The machine $M_1$ is available for not more than 6 hours 40 minutes while machine $\mathrm{M}_2$ is available for 10 hours during any working day. Formulate the problem as a LPP.
Answer
Let the firm produces x units of product A and y units of product B.
Since, each unit of product A requires one minute on machine M, and two minutes on machine $M_2$
Therefore, x units of product A will require product x minutes on machine M, and 2x minutes on machine $M_2$
Also,
Since each unit of product B requires one minute on machine M, and one minute on machine $M_2$
Therefore, y units of product A will require product y minutes on machine M, and y minutes on machine $M_2$
It is given that the machine $M_1$ is available for 6 hours and 40 minutes i.e. 400 minutes and machine $M_2$ is available for 10 hours i.e. 600 minutes.
Thus,
$\text{x}+\text{y}\leq400$
$2\text{x}+\text{y}\leq600$
Since, units of the products cannot be negative, so $\text{x},\text{y}\geq0$
Let Z denotes the total profit
$\therefore Z = 2x + 3y$ which is to be maximised
Hence, the required LPP is as follows:
Maximize $Z = 2x + 3y$
Subject to
$\text{x}+\text{y}\leq400$
$2\text{x}+\text{y}\leq600$
$\text{x},\text{y}\geq0$.
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Question 75 Marks
Maximize Z = 3x + 3y, if possible,
Subject to the constraints
$\text{x}-\text{y}\leq1$
$\text{x}+\text{y}\geq3$
$\text{x},\text{y}\geq0$
Answer
First, we will convert the given inequations into equations, we obtain the following equations:
x − y = 1, x + y = 3, x = 0 and y = 0
Region represented by x − y ≤ 1:
The line x − y = 1 meets the coordinate axes at A(1, 0) and B(0, −1) respectively.
By joining these points we obtain the line x − y = 1.
Clearly (0, 0) satisfies the inequation x + y ≤ 8.
So,the region in xy plane which contain the origin represents the solution set of the inequation x − y ≤ 1.
Region represented by x + y ≥ 3:
The line x + y = 3 meets the coordinate axes at C(3, 0) and D(0, 3) respectively.
By joining these points we obtain the line x + y = 3.
Clearly (0, 0) satisfies the inequation x + y ≥ 3.
So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + y ≥ 3.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0.
The feasible region determined by the system of constraints x − y ≤ 1, x + y ≥ 3, x ≥ 0 and y ≥ 0 are as follows.

The feasible region is unbounded.
We would obtain the maximum value at infinity.
Therefore, maximum value will be infinity i.e. the solution is unbounded.
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Question 85 Marks
A box manufacturer makes large and small boxes from a large piece of cardboard. The large boxes require 4 sq. metre per box while the small boxes require 3 sq. metre per box. The manufacturer is required to make at least three large boxes and at least twice as many small boxes as large boxes. If 60 sq. metre of cardboard is in stock, and if the profits on the large and small boxes are Rs. 3 and Rs. 2 per box, how many of each should be made in order to maximize the total profit?
Answer
Let required quantity of large and small boxes are x and y respectively.
Since, profits on each unit of large and small boxes are Rs. 3 and Rs. 2 respectively, so, profit on x units of large and y units of small boxes are Rs. 3x and Rs. 2y respectively
Let Z be total profit so,
$Z = 3x + 2y$
Since each large and small box require 4 sq. m. and 3 sq. m. cardboard respectively, so, x units of large and y units of small boxes require 4x and 3y sq.m. cardboard respectivley but only 60 sq. m. of cardboard is available, so
$4x + 3y ≤ 60$ (first constraint)
Since manufacturer is required to make at least three large boxes, so,
$X ≥ 3$ (second constraint)
Since manufacturer is required to make at least twice as many small boxes as large boxes, so,
$y ≥ 2x$ (third constraint)
Hence, mathematical formulation of LPP is find x and y which
Maximize $Z = 20x + 15y$
Subject to constriants,
$4x + 3y ≤ 60$
$x ≥ 3$
$y ≥ 2x$
$x, y ≥ 0 $ [Since production can not be less than zero]
Region $4x + 3y ≤ 60:$
Line $4x + 3y = 60$ meets axes at $A_1(15, 0), B_1(0, 20)$ respectively.
Region containing origin represents $4x + 3y ≤ 60$ as $(0, 0)$ satisfies $4x + 3y ≤ 60.$
Region $x ≥ 3:$
Line $x = 3$ is parallel to y-axis meets x-axes at $A_2(3, 0).$
Region containing origin represents $x ≥ 70$ as$(0, 0)$ satisfies $x ≥ 3.$
Region $y ≥ 2x:$
Line $y = 2x$ is passes throgh origin and $P(3, 6).$
Region containging $B_1(0, 20)$ represents $y ≥ 2x$ as $(0, 20)$ satisfies $y ≥ 2x.$
Region $x, y ≥ 0:$
It represent first quandrant.

Shaded region PQR represents feasible region.
Point $Q(6, 12)$ is obtained by solving $y = 2x$ and $4x + 3y = 60$
Point $R(3, 16)$ is obtained by solving $x = 3$ and $4x + 3y = 60.$
The value of $Z = 3x + 2y$ at
$P(3, 6) = 3(3) +2(6) = 21$
$Q(6, 12) = 3(6) + 2(12) = 42$
$R(3, 16) = 3(3) +2(16) = 41$
Maximum $Z = 42 at x = 6, y = 12$
Number of large box = 6, small box $=12$
Maximum profit - $Rs. 42.$
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Question 95 Marks
A farmer has a 100 acre farm. He can sell the tomatoes, lettuce, or radishes he can raise. The price he can obtain is Rs 1 per kilogram for tomatoes, Rs 0.75 a head for lettuce and Rs 2 per kilogram for radishes. The average yield per acre is 2000 kgs for radishes, 3000 heads of lettuce and 1000 kilograms of radishes. Fertilizer is available at Rs 0.50 per kg and the amount required per acre is 100 kgs each for tomatoes and lettuce and 50 kilograms for radishes. Labour required for sowing, cultivating and harvesting per acre is 5 man-days for tomatoes and radishes and 6 man-days for lettuce. A total of 400 man-days of labour are available at Rs 20 per man-day. Formulate this problem as a LPP to maximize the farmer's total profit.
Answer
Let the farmer sow tomatoes in x acres, lettuce in y acres & radishes in z acres of the farm.
Average yield per acre is 2000 kgs for tomatoes, 3000 kgs of lettuce and 1000 kg of radishes.
Thus, the farmer raised 2000x kg of tomatoes, 3000y kg of lettuce and 1000z kg of radishes.
Given, price he can obtain is Re 1 per kilogram for tomatoes, Re 0.75 a head for lettuce and Rs. 2 per kilogram for radishes.
$\therefore$ Selling price = Rs. [2000x(1)+3000y(0.75)+1000z(2)]
=Rs.(2000x + 2250y + 2000z)
Labour required for sowing, cultvating and harvesting per acre is 5 man-days for tomatoes and radishes and 6 man-days for lettuce.
Therefore, labour required for sowing, cultivating and harvesting per acre is 5x for tomatoes, 6y for lettuce and 5z for radishes.
Number of man-days required in sowing, cultivating and harvesting
= 5x + 6y + 5z
Price of one man-day = Rs. 20
$\therefore$ Labour cost = 20(5x + 6y + 5z) = 100x + 120y +100z
Also, fertilizer is available at Re 0.50 per kg and the amount required per acre is 100 kgs each for tomatoes and lettuce and 50 kgs for radishes.
Therefore, fertilizer required is 100x kgs for the tomatoes sown in x acres, 100y kgs for the lettuce sown in y acres and 50z kgs for radishes sown in z acres of land.
Hence, total fertilizer used= (100x + 100y +50z) kgs
Thus, fertilizer's cost
= Rs. 0.5 × (100x + 100y + 5z) = Rs. (50x + 50y + 25z)
So, the total price that has been cost to farmer = Labour cost + Fertilizer cost
= Rs. (150x + 170y + 125z)
Profit made by farmer = selling Price - cost price
= Rs. (2000x + 2250y 2000z) - Rs. (150x + 170y + 125z)
= Rs. (1850x + 2080y + 1875z)
Let Z denotes the total profit
$\therefore$ Z = 1850x + 2080y + 1875z
Now,
Total area of the farm = 100 acres
$\text{x}+\text{y}+\text{z}\leq100$
Also, it is given thet the total man - dayas available are 400.
Thus, $5\text{x}+6\text{y}+5\text{z}\leq400$
Area of the land cannot be negative.
Therefore, $\text{x}+\text{y}\geq0$
Hence, the required LPP is as follows:
Maximize Z = 1850x + 2080y + 1875z
Subject to
$\text{x}+\text{y}+\text{z}\leq100$
$5\text{x}+6\text{y}+5\text{z}\leq400$
$\text{x}+\text{y}\geq0$
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Question 105 Marks
Maximize $Z = 3x_1 + 4x_2,$ if possible,
Subject to the constraints
$\text{x}_1-\text{x}_2\leq-1$
$-\text{x}_1+\text{x}_2\leq0$
$\text{x}_1,\text{x}_2\geq0$
Answer
First, we will convert the given inequations into equations, we obtain the following equations:
$X_1 - X_2 = -1, -X_1 + x_2 = 0, X_1 = 0$ and $X_2 = 0$
Region represented by $\text{x}_1-\text{x}_2\leq-1$:
The line $x_1 - x_2 = -1$ meets the coordinate axes at A(-1, 0) and B(0, 1) respectively.
By joining these points we obtain the line $x_1 - x_2 = -1.$
Clearly (0, 0) does not satisfies the inequation $\text{x}_1-\text{x}_2\leq-1$.
So,the region in the plane which does not contain the origin represents the solution set of the inequation $\text{x}_1-\text{x}_2\leq-1$.
Region represented by $-\text{x}_1+\text{x}_2\leq0$ or $\text{x}_1\geq\text{x}_2$:
The line $-X_1 + x_2 = 0 or X_1 = x_2$ is the line passing through (0, 0).
The region to the right of the line x1 = x2 will satisfy the given inequation $-\text{x}_1+\text{x}_2\leq0$.
If we take a point (1, 3) to the left of the line $x_1 = x_2.$
Here, $1\leq3$ which is not satifying the inequation $\text{x}_1\geq\text{x}_2$.
Therefore, region to the right of the line $x_1 = x_2$ will satisfy the given inequation $-\text{x}_1+\text{x}_2\leq0$.
Region represented by $\text{x}_1\geq0$ and $\text{x}_2\geq0$:
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $\text{x}_1\geq0$ and $\text{x}_2\geq0$.
The feasible region determined by the system of constraints, $\text{x}_1-\text{x}_2\leq-1,-\text{x}_1+\text{x}_2\leq0,\text{x}_1\geq0$ and $\text{x}_2\geq0$, are as follows.
We observe that the feasible region of the given LPP does not exist.
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Question 115 Marks
Find the maximum and minimum value of 2x + y subject to the constraints:

$\text{x}+3\text{y}\geq6,\text{x}-3\text{y}\leq3,3\text{x}+4\text{y}\leq24,$ $-3\text{x}+2\text{y}\leq6,5\text{x}+\text{y}\geq5,\text{x},\text{y}\geq0$
Answer
First, we will convert the given inequations into equations, we obtain the following equations:

x + y = 4, x + y = 3, x - 2y = 2, x = 0 and y = 0.

The line x + 3y = 6 meets the coordinate axis at A(6, 0) and B(0, 2).

Join these points to obtain the line x + 3y = 6.

Clearly, (0, 0) does not satisfies the inequation $\text{x}+3\text{y}\geq6$.

So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

The line x - 3y = 3 meets the coordinate axis at C(3, 0) and D(0, -1).

Join these points to obtain the line x - 3y = 3.

Clearly, (0, 0) satisfies the inequation $\text{x}-3\text{y}\leq3$.

So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line 3x + 4y = 24 meets the coordinate axis at E(8, 0) and F(0, 6).

Join these points to obtain the line 3x + 4y = 24.

Clearly, (0, 0) satisfies the inequation $3\text{x}+4\text{y}\leq24$.

So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line -3x + 2y = 6 meets the coordinate axis at G(-2, 0) and H(0, 3).

Join these points to obtain the line -3x + 2y = 6.

Clearly, (0, 0) satisfies the inequation $-3\text{x}+2\text{y}\leq6$.

So, the region in xy-plane that contains the origin represents the solution set of the given equation.

The line 5x + y = 5 meets the coordinate axis at I(1, 0) and J(0, 5).

Join these points to obtain the line 5x + y = 5.

Clearly, (0, 0) does not satisfies the inequation $5\text{x}+\text{y}\geq5$.

So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

Region represented by $\text{x}\geq0$ and $\text{y}\geq0$:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations.

These lines are drawn using a suitable scale.
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Question 125 Marks
There are two factories located one at place P and the other at place Q. From these locations, a certain commodity is to be delivered to each of the three depots situated at A, B and C. The weekly requirements of the depots are respectively 5, 5 and 4 units of the commodity while the production capacity of the factories at P and Q are respectively 8 and 6 units. The cost of transportation per unit is given below:
How many units should be transported from each factory to each depot in order that the transportation cost is minimum. What will be the minimum transportation cost?
Answer
Here, demand of the commodity (5 + 5 + 4 = 14 units) is equal the supply of the commodity (8 + 6 = 14 units).
So, no commodity would be left at the two factories.
Let x units and y units of the commodity be transported from the factory P to the depots at A and B, respectively.

Then, (8 − x − y) units of the commodity will be transported from the factory P to the depot C.
Now, the weekly requirement of depot A is 5 units of the commodity.
Now, x units of the commodity are transported from factory P, so the remaining (5 − x) units of the commodity are transported from the factory Q to the depot A.
The weekly requirement of depot B is 5 units of the commodity.
Now, y units of the commodity are transported from factory P, so the remaining (5 − y) units of the commodity are transported from the factory Q to the depot B.
Similarly, 6 − (5 − x) − (5 − y) = (x + y − 4) units of the commodity will be transported from the factory Q to the depot C.
Since the number of units of commodity transported are from the factories to the depots are non-negative, therefore,

x ≥ 0, y ≥ 0, 8 − x − y ≥ 0, 5 − x ≥ 0, 5 − y ≥ 0, x + y − 4 ≥ 0

Or x ≥ 0, y ≥ 0, x + y ≤ 8, x ≤ 5, y ≤ 5, x + y ≥ 4

Total transportation cost = 160x + 100y + 150(8 − x − y) + 100(5 − x) + 120(5 − y) + 100(x + y − 4) = 10x − 70y + 1900

Thus, the given linear programming problem is
Minimise Z = 10x − 70y + 1900
Subject to the constraints
x + y ≤ 8
x ≤ 5
y ≤ 5
x + y ≥ 4
x ≥ 0, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented as,

The coordinates of the corner points of the feasible region are A(4, 0), B(5, 0), C(5, 3), D(3, 5), E(0, 5) and F(0, 4).

The value of the objective function at these points are given in the following table.
Corner Point
Z = 10x - 70y + 1900
(4, 0)
10 × 4 - 70 × 0 + 1900 = 1940
(5, 0)
10 × 5 - 70 × 0 + 1900 = 1950
(5, 3)
10 × 5 - 70 × 3 + 1900 = 1740
(3, 5)
10 × 3 - 70 × 5 + 1900 = 15800
(0, 5)
10 × 0 - 70 × 5+ 1900 = 1550 → Maximum
(0, 4) 10 × 0 - 70 × 4 + 1900 = 1620
The minimum value of Z is 1550 at x = 0, y = 5.
Hence, for minimum transportation cost, factory P should supply 0, 5, 3 units of commodity to depots A, B, C respectively and factory Q should supply 5, 0, 1 units of commodity to depots A, B, C respectively.
The minimum transportation cost is Rs. 1,550.
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Question 135 Marks
Vitamins $A$ and $B$ are found in two different foods $F_1$ and $F_2$. One unit of food $F_1$ contains 2 units of vitamin $A$ and 3 units of vitamin $B$. One unit of food $F_2$ contains 4 units of vitamin $A$ and 2 units of vitamin $B$. One unit of food $F_1$ and $F_2$ cost Rs 50 and 25 respectively. The minimum daily requirements for a person of vitamin $A$ and $B$ is 40 and 50 units respectively. Assuming that anything in excess of daily minimum requirement of vitamin $A$ and $B$ is not harmful, find out the optimum mixture of food $\mathrm{F}_1$ and $\mathrm{F}_2$ at the minimum cost which meets the daily minimum requirement of vitamin A and B . Formulate this as a LPP.
Answer
Let x and y units of food $F_1$ and food $F_2$ were mixed.
Clearly, $\text{x}\geq0$ and $\text{y}\geq0$
One unit of food $F_1$ contains 2 units of vitamin A and one unit of of food $F_2$ contains 4 units of vitamin A.
Therefore, x and y units of food $F_1$ and food $F_2$ respectively contains 2x and 4y units of vitamin A.
It is given that the minimum daily requirements for a person of vitamin A is 40 units.
Hence, $2\text{x}+4\text{y}\geq40$
One unit of food $F_1$ contains 3 units of vitamin B and one unit of food $F_2$ contains 2 units of of vitamin B.
Therefore, x and y units of $F_1$ and $F_2$ respectively contains 3x and 2y units of vitamin B.
It is given that the minimum daily requirements for a person of vitamin B is 50 units.
​​​​​​​Hence, $3\text{x}+2\text{y}\geq50$
One unit of food $F_1$ and food $F_2$ cost Rs 50 and 25 respectively.
Therefore, x and y units of food $F_1$ and food $F_2$ costs Rs. 50x and Rs. 25y respectively.
Let Z denote the total cost
Then, $Z = Rs. (50x + 25y)$
Hence, the required LPP is Minimize $Z = 50x + 25y$
subject to
$2\text{x}+4\text{y}\geq40$
$3\text{x}+2\text{y}\geq50$
$\text{x}\geq0,\text{y}\geq0$
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Question 145 Marks
A publisher sells a hard cover edition of a text book for Rs. 72.00 and paperback edition of the same ext for Rs. 40.00. Costs to the publisher are Rs. 56.00 and Rs. 28.00 per book respectively in addition to weekly costs of Rs. 9600.00. Both types require 5 minutes of printing time, although hardcover requires 10 minutes binding time and the paperback requires only 2 minutes. Both the printing and binding operations have 4,800 minutes available each week. How many of each type of book should be produced in order to maximize profit?
Answer
Let the sale of hand cover edition be 'h' and that of paperback editions be 't'.
SP of a hard cover edition of the textbook $= Rs. 72$
SP of a paperback edition of the textbook $= Rs. 40$
Cost to the publisher for hard cover edition $= Rs. 56$
Cost to the publisher for a paperback edition $= Rs. 28$
Weekly cost to the publisher = Rs. 9600
Profit to be maximised, $Z = (72 − 56) h + (40 − 28) t − 9600$
$⇒ Z = 16h + 12t – 9600$
$5(h + t) ≤ 4800$
$10h + 2t ≤ 4800$

The corner points are $O(0, 0), B_1(0, 960), E_1(360, 600)$ and $F_1(480, 0).$
The values of Z at these corner points are as follows:
Corner point
$Z = 16h + 12t - 9600$
O
$-9600$
$B_1$
 $1920$
$G_1$
 $3360$
$F_1$
 -$1920$
The maximum value of Z is $3360$ which is attained at $E_1(360, 600).$
The maximum profit is $3360$ which is obtained by selling 360 copies of hardcover edition and 600 copies of paperback edition.
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Question 155 Marks
Amit's mathematics teacher has given him three very long lists of problems with the instruction to submit not more than 100 of them (correctly solved) for credit. The problem in the first set are worth 5 points each, those in the second set are worth 4 points each, and those in the third set are worth 6 points each. Amit knows from experience that he requires on the average 3 minutes to solve a 5 point problem, 2 minutes to solve a 4 point problem, and 4 minutes to solve a 6 point problem. Because he has other subjects to worry about, he can not afford to devote more than $3\frac{1}{2}$ hours altogether to his mathematics assignment. Moreover, the first two sets of problems involve numerical calculations and he knows that he cannot stand more than $2\frac{1}{2}$ hours work on this type of problem. Under these circumstances, how many problems in each of these categories shall he do in order to get maximum possible credit for his efforts? Formulate this as a LPP.
Answer
Given information can be tabulated as below:
Sets
Time requirement
points
I
3
5
II
2
 
III
4
6
Time for all three sets $=3\frac{1}{2}$ hours
Time for set I and II $=2\frac{1}{2}$ hours
Number of quations maximum 100
Given, each question from set I, II, III earn 5,4,6 points respectively, so x questions of set I, y questions of set II and z questions of set III earn 5x, 4y and 6z points, let total point credit be U

So, U = 5x + 4y + 6z

Given, each question of set I, II and III require 3,2 and 4 minutes respectively, so x questions of set I, y questions of set II and z questions of set III require 3x, 2y and 4z mimutes respectively but given that total time to devote in all three sets is

$3\frac{1}{2}$ hours = 210 minutes and first two sets is $2\frac{1}{2}$ hours = 150 minutes

So,

$3\text{x}+2\text{y}+4\text{z}\leq210$ (First constraint)

$3\text{x}+2\text{y}\leq150$ (Second constraint)

Given, total number of questions cannot exceed 100

So, $\text{x}+\text{y}+\text{z}\leq100$ (Third constraint)

Hence, mathematical formulation of LPP is

Find x and y which maximize U = 5x + 4y + 6z

Subject to constraint,

$3\text{x}+2\text{y}+4\text{z}\leq210$

$3\text{x}+2\text{y}\leq150$

$\text{x}+\text{y}+\text{z}\leq100$
$\text{x},\text{y},\text{z}\geq0$

[Since number of questions to solve from each set cannot be less than zero].
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Question 165 Marks
A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of first machine is 12 hours and that of second machine is 9 hours per day. Each unit of product A requires 3 hours on both machines and each unit of product B requires 2 hours on first machine and 1 hour on second machine. Each unit of product A is sold at Rs. 7 profit and that of B at a profit of Rs. 4. Find the production level per day for maximum profit graphically.
Answer
Let x units of product A and y units of product B be manufactured by the manufacturer per day.
It is given that one unit of product A requires 3 hours of processing time on first machine, while one unit of product B requires 2 hours of processing time on first machine.
It is also given that first machine is available for 12 hours per day.
$\therefore$ 3x + 2y ≤ 12
Also, one unit of product A requires 3 hours of processing time on second machine, while one unit of product B requires 1 hour of processing time on second machine.
It is also given that second machine is available for 9 hours per day.
$\therefore$ 3x + y ≤ 9
The profits on one unit each of product A and product B is Rs. 7 and Rs 4, respectively.
So, the objective function is given by Z = Rs. (7x + 4y).
Therefore, the mathematical formulation of the given linear programming problem can be stated as:

Maximize Z = 7x + 4y
Subject to the constraints
3x + 2y ≤ 12 .....(1)
3x + y ≤ 9 .....(2)
x ≥ 0, y ≥ 0 .....(3)
The feasible region determined by constraints (1) to (3) is graphically represented as:

Here, it is seen that OABCO is the feasible region and it is bounded.
The values of Z at the corner points of the feasible region are represented in tabular form as:
Corner point
Z = 7x + 4y
O(0, 0)
Z = 7 × 0 + 4 × 0 = 0
A(0, 10)
Z = 7 × 3 + 4 × 0 = 21
B(173, 0)
Z = 7 × 2 + 4 × 3 = 26
C(3, 8)
Z = 7 × 0 + 4 × 6 = 24
The maximum value of Z is 26, which is obtained at x = 2 and y = 3.
Thus, 2 units of product A and 3 units of product B should be manufactured by the manufacturer per day in order to maximize the profit.
Also, the maximum daily profit of the manufacturer is Rs. 26.
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Question 175 Marks
Maximum Z = 4x + 3y
Subject to
$3\text{x}+4\text{y}\leq24$
$8\text{x}+6\text{y}\leq48$
$\text{x}\leq5$
$\text{y}\leq6$
$\text{x},\text{y}\geq0$
Answer
We need to maximize Z= 4x + 3y
First, we will convert the given inequations into equations, we obtain the following equations:
3x + 4y = 24, 8x + 6y = 48, x = 5, y = 6, x = 0 and y = 0.
The line 3x + 4y = 24 meets the coordinate axis at A(8, 0) and B(0, 6).
Join these points to obtain the line 3x + 4y = 24.
Clearly, (0, 0) satisfies the inequation $3\text{x}+4\text{y}\leq24$.
So, the region in xy-plane that contains the origin represents the solution set of the given equation.
The line 8x + 6y = 48 meets the coordinate axis at C(6, 0) and D(0, 8).
Join these points to obtain the line 8x + 6y = 48.
Clearly, (0, 0) satisfies the inequation $8\text{x}+6\text{y}\leq48$.
So, the region in xy-plane that contains the origin represents the solution set of the given equation.
x = 5 is the line passing through x = 5 parallel to the Y axis.
y = 6 is the line passing through y = 6 parallel to the X axis.
Region represented by $\text{x}\geq0$ and $\text{y}\geq0$: Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.

The corner points of the feasible region are O(0, 0), G(5, 0), $\text{F}\Big(5,\frac{4}{3}\Big),\text{E}\Big(\frac{24}{7},\frac{24}{7}\Big)$ and B(0, 6).
The values of Z at these corner points are as follows.
$\text{Corner point}$ $\text{Z}=4\text{x}+3\text{y}$
$\text{O}(0, 0)$ $4\times0+3\times0=0$
$\text{G}(5, 0)$ $4\times5+3\times0=20$
$\text{F}\Big(5,\frac{4}{3}\Big)$ $4\times5+3\times\frac{4}{3}=24$
$\text{E}\Big(\frac{24}{7},\frac{24}{7}\Big)$ $4\times\frac{24}{7}+3\times\frac{24}{7}=\frac{196}{7}=24$
$\text{B}(0, 6)$ $4\times0+3\times6=18$
We see that the maximum value of the objective function Z is 24 which is at $\text{F}\Big(5,\frac{4}{3}\Big)$ and $\text{E}\Big(\frac{24}{7},\frac{24}{7}\Big)$.
Thus. the optimal value of is 24.
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Question 185 Marks
A small manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry, then sent to the machine shop for finishing. The number of man-hours of labour required in each shop for the production of each unit of A and B, and the number of man-hours the firm has available per week are as follows:
Gadget
Fondry
Machine-shop
A
B
10
6
5
4
Firm's capacity per week
1000
600
The profit on the sale of A is Rs. 30 per unit as compared with Rs. 20 per unit of B. The problem is to determine the weekly production of gadgets A and B, so that the total profit is maximized. Formulate this problem as a LPP.
Answer
The given data may be put in the following tabular from:-
Gadget
Fondry
Machine-shop
Profit
A
B
10
6
5
4
Rs. 30
Rs. 20
Firm's capacity per week
1000
600
 
Let required weekly production of gadgets A and B be x and y respectively.
Given that, profit on each gadget A is Rs 30
So, profit on x gadget of type A = 30x
Profit on each gadget of type B = Rs. 20
So, profit on y gadget of type B = 20y
Let Z denote the total profit, so
Z = 30x + 20y
Given, production of one gadget A requires 10 hours per week for foundry and gadget B requires 6 hours per week for foundry.
So, x units of gadget A requires 10x hours per week and y units of gadget B requires by hours per week, But the maximum capacity of foundry per week is 1000 hours, so
10x + 6y s 1000
This is first constraint.
Given, production of one unit gadget A requires 5 hours per week of machine shop and production of one unit of gadget B requires 4 hours per week of machine shop.
So, x units of gadget A requires 5x hours per week and y units of gadget B requires 4y hours per week, but the maximum capacity of machine shop is 600
hours per week.
So, 5x + 4y = 600
This is second constraint.
Hence, mathematical formulation of LPP is:
Find x and y which
Maximize Z = 30x + 2y
Subject to constraints,
10x + 6y ≤ 1000
5x + 4y ≤ 600
And, x, y ≥ 0 [Since production cannot be less than zero]
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Question 195 Marks
A library has to accommodate two different types of books on a shelf. The books are 6 cm and 4 cm thick and weigh 1kg and $1\frac{1}{2}$ kg each respectively. The shelf is 96 cm long and at most can support a weight of 21kg. How should the shelf be filled with the books of two types in order to include the greatest number of books? Make it as an LPP and solve it graphically.
Answer
Let x books of first type and y books of second type were accommodated. Number of books cannot be negative.

Therefore, x, y ≥ 0

According to question, the given information can be tabulated as:
  Thickness(cm) Weight(kg)
First type(x) 6 1
Second type(v) 4 1.5
Capacity of shelf 96 21
Therefore, the constraints

6x + 4y ≤ 96

x + 1.5y ≤ 21

Number of books = Z = x + y which is to be maximised

Thus, the mathematical formulation of the given linear programmimg problem is

Max Imize Z = x + y

Subject to

6x + 4y ≤ 96

x + 1.5y ≤ 21

First we will convert inequations into equations as follows:

6x + 4y = 96, x + 1.5y = 21, x = 0 and y = 0

Region represented by 6x + 4y ≤ 96:

The line 6x + 4y = 96 meets the coordinate axes at A(16, 0) and B(O, 24) respectively.

By joining these points we obtain the line 6x + 4y = 96.

Clearly (0, 0) satisfies the 6x + 4y = 96.

So, the region which contains the origin represents the solution set of the inequation 6x + 4y ≤ 96.

Region represented by x + 1.5y ≤ 21:

The line x + 1.5y = 21 meets the coordinate axes at C(21, 0) and D(O, 14) respectively.

By joining these points we obtain the line x + 1.5y = 21.

Clearly (0, 0) satisfies the inequation x + 1.5y ≤ 21.

So, the region which contains the origin represents the solution set of the inequation x + 1.5y ≤ 21.

Region represented by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.

The feasible region determined by the system of constraints 6x + 4y ≤ 96, x + 1.5y ≤ 21, x ≥ 0 and y ≥ 0 are as follows.



The corner points are O(0, 0), D(0, 14), E(12, 6), A(16,0)

The values of Z at these corner points are as follows.
Corner point
Z = x + y
O
0
D
14
E
18
A
16
The maximum value of Z is 18 which is attained at E(12, 6).

Thus, maximum number of books that can be arranged on shelf is 18 where 12 books are of first type and 6 books are the other type.
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Question 205 Marks
A company produces two types of leather belts, say type A and B. Belt A is a superior quality and belt B is of a lower quality. Profits on each type of belt are Rs. 2 and Rs. 1.50 per belt, respectively. Each belt of type A requires twice as much time as required by a belt of type B. If all belts were of type B, the company could produce 1000 belts per day. But the supply of leather is sufficient only for 800 belts per day (both A and B combined). Belt A requires a fancy buckle and only 400 fancy buckles are available for this per day. For belt of type B, only 700 buckles are available per day.
How should the company manufacture the two types of belts in order to have a maximum overall profit?
Answer
Let the company produces x belts of type A and y belts of type B.
Number of belts cannot be negative.
Therefore, $\text{x},\text{y}\geq0$
It is given that leather is sufficient only for 800 belts per day (both A and B combined).
Therefore, $\text{x}+\text{y}\leq800$
It is given that the rate of production of belts of type B is 1000 per day.
Hence, the time taken to produce y belts of type B is $\frac{\text{y}}{1000}$
And, since each belt of type A requires twice as much time as a belt of type B, the rate of production of belts of type A is 500 per day and therefore, total time taken to produce x belts of type A is *
Thus, we have
$\frac{\text{x}}{500},\frac{\text{y}}{1000}\leq1$
$\Rightarrow2\text{x}+\text{y}\leq1000$
Belt A requires a fancy buckle and only 400 fancy buckles are available for this per day.
$\text{x}\leq400$
For belt of type B, only 700 buckles are available per day.
$\text{y}\leq700$
Profits on each type of belt are Rs. 2 and Rs. 1.50 per belt, respectively.
Therefore, profit gained on x belts of type A and y belts of type B is Rs. 2x and Rs. 1.50 yrespectively.
Hence, the total profit would be Rs. $(2x + 1.50y).$
Let Z denote the total profit.
$Z = 2x + 1.5y$
Thus, the mathematical formulation of the given linear programming problem is Max $Z = 2x + 1.5y$
Subject to
Max $Z = 2x + 1.5y$
Subject to
$\text{x}+\text{y}\leq800$
$2\text{x}+\text{y}\leq1000$
$\text{x}\leq400$
$\text{y}\leq700$
$\text{x},\text{y}\geq0$
First we will convert inequations into equations as follows:
$x + y = 800$
$2x + y = 1000$
$x = 400$
$y = 700$
$x = 0$
$y = 0$
Region represented by $\text{x}+\text{y}\leq800$:
The line $x + y = 800$ meets the coordinate axes at $A_1(800, 0)$ and $B_1(0, 800)$ respectively.
By joining these points we obtain the line $x + y = 800.$
Clearly (0, 0) satisfies the $x + y = 800.$
So, the region which contains the origin represents the solution set of the inequation $\text{x}+\text{y}\leq800$.
Region represented by $2\text{x}+\text{y}\leq1000$:
The line $2x + y = 1000$ meets the coordinate axes at $C_1(500, 0)$ and $D_1(0, 1000)$ respectively.
By joining these points we obtain the line $2x + y = 1000.$
Clearly (0, 0) satisfies the inequation $2\text{x}+\text{y}\leq1000$.
So,the region which contains the origin represents the solution set of the inequation $2\text{x}+\text{y}\leq1000$
Region represented by $\text{x}\leq400$:
The line $x = 400$ will pass through $E_1(400, 0).$
The region to the left of the line x = 400 will satisfy the inequation $\text{x}\leq400$.
Region represented by $\text{y}\leq700$:
The line $y = 700$ will pass through $F_1(0, 700).$
The region below the line $\text{y}\leq700$ will satisfy the inequation $\text{y}\leq700$.
Region represented by $\text{x}\geq0$ and $\text{y}\geq0$:
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $\text{x}\geq0$, and $\text{y}\geq0$.
The feasible region determined by the system of constraints $\text{x}+\text{y}\leq800,2\text{x}+\text{y}\leq1000,\text{x}\leq400,\text{y}\leq700,\text{x}\geq0$ and $\text{y}\geq0$ are as follows.
The feasible region determined by the system of constraints is:

The corner points are $F_1(0, 700), G_1(200, 600), H_1(400, 200)$ and $E_1(400, 0).$
The values of Z at these corner points are as follows.
Corner point
Z = 2x + 1.5y
$F_1(0, 700)$
 1050
$G_1(200, 600)$
 1300
$H_1(400, 200)$
 1100
$E_1(400, 0)$ 800
The maximum value of Z is 1300 which is attained at $G_1(200, 600).$
Thus, the maximum profit is Rs. 1300 obtained when 200 belts of type A and 600 belts of type 8 were produced.
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Question 215 Marks
Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contains at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and food Q costs Rs 80/kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.
Answer
Let x units of food P and y units of food Q are mixed together to make the mixture.

The cost of food P is Rs. 60/kg and that of Q is Rs. 80/kg.

So, x kg of food P and ykg of food Q will cost Rs. (60x+ 80y).

Since one kg of food P contains 3 units of vitamin A and one kg of food Q contains 4 units of vitamin A, therefore, x kg of food P and y kg of food Q will contain (3x+4y) units of vitamin A.

But, the mixture should contain atleast 8 units of vitamin A.

$3\text{x}+4\text{y}\geq8$

Similarly, x kg of food and y kg of food Q will contain (5x + 2y) units of vitamin B.

But, the mixture should contain atleast 11 units of vitamin B. 5x+ 2y 11.

Thus, the given linear programming problem is Minimise Z = 60x + 80y

Subject to the constraints

$3\text{x}+4\text{y}\geq8$

$5\text{x}+2\text{y}\geq11$

$\text{x},\text{y}\geq0$

The feasible region determined by the given constraints can be diagrammatically represented as,



The coordinates of the corner points of the feasible region are points of the feasible region are $\text{A}\Big(\frac{8}{3},0\Big)\text{B}\Big(2,\frac{1}{2}\Big)$ and $\text{C}\Big(0,\frac{11}{2}\Big)$.

The value of the objective function at these points are given in the following table.
$\text{Corner point}$ $\text{Z}=60\text{x}+80\text{y}$
$\Big(\frac{8}{3},0\Big)$ $60\times\frac{8}{3}+80\times0=160\rightarrow\text{Minimum}$
$\Big(2,\frac{1}{2}\Big)$ $60\times2+80\times\frac{1}{2}=160\rightarrow\text{Minimum}$
$\Big(0,\frac{11}{2}\Big)$ $60\times0+80\times\frac{11}{2}=440$
The smallest value of Z is 160 which is obtained at the points $\Big(\frac{8}{3},0\Big)$ and $\Big(2,\frac{1}{2}\Big)$.

It can be verified that the open half-plane represented by $60\text{x}+80\text{y}\leq160$ has no common points with the feasible region.

So, the minimum value of Z is 160.

Hence, the minimum cost of the mixture is Rs. 160.
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Question 225 Marks
Maximum Z = x - 5y + 20
Subject to
$\text{x}-\text{y}\geq0$
$-\text{x}+2\text{y}\geq2$
$\text{x}\geq3$
$\text{y}\geq4$
$\text{x},\text{y}\geq0$
Answer
First, we will convert the given inequations into equations, we obtain the following equations:

x - y = 0, -x + 2y = 2, x = 3, y = 4, x = 0 and y = 0.

Region represented by $\text{x}-\text{y}\geq0$ or $\text{x}\geq\text{y}$:

The line x - y = 0 or x = y passes through the origin.

The region to the right of the line x = y will satisfy the given inequation.

Let's check by taking an example like if we take a point (4, 3) to the right of the line x = y.

Here $\text{x}\geq\text{y}$.

So, it satisfy the given inequation.

Take a point (4, 5) to the left of the line x = y.

Here, $\text{x}\leq\text{y}$.

That means it does not satisfy the given inequation.

Region represented by $-\text{x}+2\text{y}\geq2$:

The line -x + 2y = 2 meets the coordinate axes at A(-2, 0) and B(0, 1) respectively.

By joining these points we obtain the line - x + 2y = 2.

Clearly (0, 0) does not satisfies the inequation $-\text{x}+2\text{y}\geq2$.

So, the region in xy plane which does not contain the origin represents the solution set of the inequation $-\text{x}+2\text{y}\geq2$.

The line x = 3 is the line that passes through the point (3, 0) and is parallel to Y axis. $\text{x}\geq3$ is the region to the right of the line x = 3.

The line y = 4 is the line that passes through the point (0, 4) and is parallel to X axis. $\text{y}\geq4$ is the region below the line y = 4.

Region represented by $\text{x}\geq0$ and $\text{y}\geq0$:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations $\text{x}\geq0$ and $\text{y}\geq0$.

The feasible region determined by the system of constraints $\text{x}-\text{y}\geq0,-\text{x}+2\text{y}\geq2,\text{x}\geq3,\text{y}\geq4,\text{x}\geq0,$and $\text{y}\geq0$ are as follows.



The corner points of the feasible region are $\text{C}\Big(3,\frac{5}{2}\Big),\text{D}(3, 3),\text{E}(4, 4)$ and $\text{F}(6, 4)$.

The values of Z at these corner points are as follows.
$\text{Corner point}$
$\text{Z}=\text{x}-5\text{y}+20$
$\text{C}\Big(3,\frac{5}{2}\Big)$
$3-5\times\frac{5}{2}+20=\frac{21}{2}$
$\text{D}(3, 3)$
$3-5\times3+20=8$
$\text{E}(4, 4)$
$4-5\times4+20=4$
$\text{F}(6, 4)$
$6-5\times4+20=6$
Therefore, the minimum value of Z is 4 at the point E(4, 4).

Hence, x = 4 and y = 4 is the optimal solution of the given LPP.

Thus, the optimal value of Z is 4.
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Question 235 Marks
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman's time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman's time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman's time.
  1. What number of rackets and bats must be made if the factory is to work at full capacity?
  2. If the profit on a racket and on a bat is Rs. 20 and Rs. 10 respectively, find the maximum profit of the factory when it works at full capacity.
Answer
Let x number of tennis rackets and y number of cricket bats were sold.
Number of tennis rackets and cricket balls cannot be negative.
Therefore, $x ≥ 0, y ≥ 0$
It is given that a tennis racket takes 1.5 hours of machine time and 3 hours of craftman's time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman's time.
Also, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman's time.
Therefore,
1.5 x plus 3 y less or equal than 42
3 x plus y less or equal than 24
If the profit on a racket and on a bat is Rs. 20 and Rs. 10 respectively.
Therefore, profit made on x tennis rackets and y cricket bats is Rs. 20x and Rs. 10y respectively.
Total profit = $Z = 20x + 10y$
The mathematical form of the given LPP is:
Maximize $Z = 20x + 10y$
Subject to constraints:
$1.5$ x plus 3 y less or equal than $42$
$3$ x plus y less or equal than $24$
$x ≥ 0, y ≥ 0$
First we will convert inequations into equations as follows:
$1.5x + 3y = 42, 3x + y = 24, x = 0$ and $y = 0$
Region represented by $1.5x + 3y ≤ 42:$
The line $1.5x + 3y = 42$ meets the coordinate axes at $A_1(28, 0)$ and $B_1(0, 14)$ respectively.
By joining these points we obtain the line $1.5x + 3y = 42.$
Clearly (0, 0) satisfies the $1.5x + 3y = 42.$
So, the region which contains the origin represents the solution set of the inequation $1.5x + 3y ≤ 42.$
Region represented by $3x + y ≤ 24:$
The line $3x + y = 24$ meets the coordinate axes at $C_1(8,0)$ and $D_1(0, 24)$ respectively.
By joining these points we obtain the line $3x + y = 24.$
Clearly (0, 0) satisfies the inequation $3x + y ≤ 24$.
So the region which contains the origin represents the solution set of the inequation $3x + y ≤ 24.$
Region represented by $x ≥ 0$ and $y ≥ 0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $x ≥ 0$ and $y ≥ 0.$
The feasible region determined by the system of constraints $1.5x + 3y ≤ 42, 3x + y ≤ 24, x ≥ 0$ and $y ≥ 0$ are as follows.

In the above graph, the shaded region is the feasible region.
The corner points are $O(0, 0), B_1(0, 14), E_1(04, 12),$ and $C_1(8, 0).$
The values of the objective function Z at corner points of the feasible region are given in the following table:
Corner Points
Z = 20x + 10y
 
$O(0, 0)$
$0$
 
$B_1(0, 14)$
$140$
 
$E_1(4, 12)$
$200$
Maximum
$C_1(8, 0)$
$160$
  
Clearly, Z is maximum at x = 4 and y= 12 and the maximum value of Z at this point is 200.
Thus, maximum profit is of Rs. 200 obtained when 4 tennis rackets and 12 cricket bats were sold.
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Question 245 Marks
A manufacturing company makes two models A and B of a product. Each piece of model A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each piece of model B requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of Rs. 8000 on each piece of model A and Rs. 12000 on each piece of model B. How many pieces of model A and model B should be manufactured per week to realise a maximum profit? What is the maximum profit per week?
Answer
The given data can be written in the tabular form as follows:
Model
A
B
Maximum hours
Fabricating
9
12
180
Finishing
1
3
30
Profit
8000
12000
  
Let x be the number of pieces of A and y be the number of pieces of B manufactured to earn the maximum profit.
Then the mathematical model of the LPP is as follows:
Maximize $Z = 8000x + 12000$ Subject to $9x + 12y ≤ 180, x + 3y ≤ 30$ and $x ≥ 0, y ≥ 0.$
To solve the LPP We draw the lines, $9x + 12y = 180, x + 3y = 30$
The feasible region of the LPP is shaded in graph.

The coordinates of the vertices (Corner - points) of shaded feasible region ABC are $A (20, 0), B(12,6)$ and $C(0, 10).$
The values of the objective of function at these points are given in the following table:
Paint $(X_1, X_1)$
Value of objective function Z = 8,000x + 12,000y
$A(20, 0)$
$Z = 1,60,000$
$B(12, 6)$
$Z = 1,68,000$
$C(0, 10)$
$Z = 1,20,000$
12 pieces of Model A and 6 pieces of Model B should be eaned maximize the profit.
The maximum profit that can be eared is Rs. $1,68,000.$
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Question 255 Marks
A diet of two foods $F_1$ and $F_2$ contains nutrients thiamine, phosphorous and iron.
The amount of each nutrient in each of the food (in milligrams per 25gms) is given in the following table:
Nutrients Food $F_1$ $F_2$
Thiamine 0.25 0.10
Phosphorous 0.75 1.50
Iron 1.60 0.80
The minimum requirement of the nutrients in the diet are 1.00mg of thiamine, 7.50mg of phosphorous and 10.00mg of iron.
The cost of $F_1$ is 20 paise per 25gms while the cost of $F_2$ is 15 paise per 25gms.
Find the minimum cost of diet.
Answer
Let 25x grams of food $F_1$ and 25y grams of food $F_2$ be used to fulfil the minimum requirement of thiamine, phosphorus and iron.
As, we are given,
Nutrients Food $F_1$ $F_2$
Thiamine 0.25 0.10
Phosphorous 0.75 1.50
Iron 1.60 0.80
And the minimum requirement of the nutrients in the diet are 1.00mg of thiamine, 7.50mg of phosphorous and 10.00mg of iron.
Therefore, $0.25\text{x}+0.10\text{y}\geq1$
$0.75\text{x}+1.50\text{y}\geq7.5$
$1.6\text{x}+0.8\text{y}\geq10$
Since the quantity cannot be negative
$\therefore\text{x},\text{y}\geq0$
The cost of $F_1$ is 20 paise per 25gms while the cost of $F_2$ is 15 paise per 25 gms.
Therefore, the cost of 25x grams of food $F_1$ and 25y grams of food $F_2$ is Rs. $(0.20x + 0.15y).$
Hence,
Minimize $Z = 0.20x + 0.15y$
Subject to
$0.25\text{x}+0.10\text{y}\geq1,0.75\text{x}+1.50\text{y}\geq7.5,$ $1.6\text{x}+0.8\text{y}\geq10,\text{x},\text{y}\geq0.$
First, we will convert the given inequations into equations, we obtain the following equations:
$0.25x + 0.10y = 1, 0.75x + 1.50y = 7.5, 1.6x + 0.8y = 10, x = 0$ and $y = 0.$
The line $0.25x + 0.10y = 1$ meets the coordinate axis at A(4, 0) and B(0, 10).
Join these points to obtain the line $0.25x + 0.10y = 1.$
Clearly, (0, 0) does not satisfies the inequation $0.25\text{x}+0.10\text{y}\geq1$.
So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
The line $0.75x + 1.50y = 7.5.$ meets the coordinate axis at $C(10, 0)$ and $D(0, 5)$.
Join these points to obtain the line $0.75x + 1.50y = 7.5.$
Clearly, (0, 0) does not satisfies the inequation $0.75\text{x}+1.50\text{y}\geq7.5$.
So, the region in xyplane that does not contains the origin represents the solution set of the given equation.
The line $1.6x + 0.8y = 10$ meets the coordinate axis at $\text{E}\Big(\frac{25}{4},0\Big)$ and $\text{F}\Big(0,\frac{25}{2}\Big).$
Join these points to obtain the line $1.6x + 0.8y = 10.$
Clearly, (0, 0) does not satisfies the inequation $1.6\text{x}+0.8\text{y}\geq10$.
So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
Region represented by $\text{x}\geq0$ and $\text{y}\geq0$:
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.

The corner points of the feasible region are $F(0, 12.5), G(5, 2.5), C(10, 0)$
The value of the objective function at these points are given by the following table:
Points Value of Z
F $0.20(0) + 0.15(12.5) = 1.875$
G $0.20(5) + 0.15(2.5) = 1.375$
C $0.20(10) + 0.15(0) = 200$
Thus, the minimum cost is at G which is Rs. 1.375.
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Question 265 Marks
A factory manufactures two types of screws, A and B, each type requiring the use of two machines - an automatic and a hand-operated. It takes 4 minute on the automatic and 6 minutes on the hand-operated machines to manufacture a package of screws 'A', while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a package of screws 'B'. Each machine is available for at most 4 hours on any day. The manufacturer can sell a package of screws 'A' at a profit of 70 P and screws 'B' at a profit of Rs. 1. Assuming that he can sell all the screws he can manufacture, how many packages of each type should the factory owner produce in a day in order to maximize his profit? Determine the maximum profit.
Answer
Let the factory manufacture x screws of type A and y screws of type B on each day.

Therefore, $\text{x}\geq0$ and $\text{y}\geq0$

The given information can be compiled in a table as follows.
 
Screw A
Screw B
Availability
Automatic Machine (min)
4
6
4 × 60 = 120
Hand Operated Machine (min)
6
3
4 × 60 = 120
The profit on a package of screws A is Rs. 7 and on the package of screws B is Rs. 10.

Therefore, the constraints are

$4\text{x}+6\text{y}\geq240$

$6\text{x}+3\text{y}\geq240$

Total profit, Z = 7x + 10y.

The mathematical formulation of the given problem is,

Maximize Z = 7x + 10y ...(1)

Subject to the constraints, $4\text{x}+6\text{y}\geq240\dots(2)$

$6\text{x}+3\text{y}\geq240\dots(3)$

$\text{x},\text{y}\geq0\dots(4)$

The feasible region determined by the system of constraints is



The corner points are A(40, 0), B(30, 20) and C(0, 40).

The values of Z at these corner points are as follows.
Corner point
Z = 7x + 10y
 
A(40, 0)
280
 
B(30, 20)
410
→ Maximum
C(0, 40)
400
  
The maximum value of Z is 410 at (30, 20).

Thus, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs 410.
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Question 275 Marks
Maximize Z = 18x + 10y
Subject to
$4\text{x}+\text{y}\geq20$
$2\text{x}+3\text{y}\geq30$
$\text{x},\text{y}\geq0$
Answer
First, we will convert the given inequations into equations, we obtain the following equations: 4x + y = 20, 2x + 3y = 30, x = 0 and y = 0 Region represented by 4x + y ≥ 20: The line 4x + y = 20 meets the coordinate axes at A(5, 0) and B(0, 20) respectively. By joining these points we obtain the line 4x + y = 20. Clearly (0, 0) does not satisfies the inequation 4x + y ≥ 20. So, the region in xy plane which does not contain the origin represents the solution set of the inequation 4x + y ≥ 20. Region represented by 2x + 3y ≥ 30: The line 2x + 3y = 30 meets the coordinate axes at C(15, 0) and D(0, 10) respectively. By joining these points we obtain the line 2x + 3y = 30. Clearly (0, 0) does not satisfies the inequation 2x + 3y ≥ 30. So, the region which does not contain the origin represents the solution set of the inequation 2x + 3y ≥ 30. Region represented by x ≥ 0 and y ≥ 0: Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0. The feasible region determined by the system of constraints, 4x + y ≥ 20, 2x + 3y ≥ 30, x ≥ 0, and y ≥ 0, are as follows.
The corner points of the feasible region are B(0, 20), C(15, 0), E(3, 8) and C(15, 0). The values of Z at these corner points are as follows.
Corner point
Z = 18x + 10y
B(0, 20)
18 × 0 + 10 × 20 = 200
E(3, 8)
18 × 3 + 10 × 8 = 134
C(15, 0)
18 × 15 + 10 × 0 = 270
Therefore, the minimum value of Z is 134 at the point E(3, 8). Hence, x = 3 and y = 8 is the optimal solution of the given LPP. Thus, the optimal value of Z is 134.
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Question 285 Marks
A firm manufactures two types of products $A$ and $B$ and sells them at a profit of Rs. 5 per unit of type $A$ and Rs 3 per unit of type $B$. Each product is processed on two machines $M_1$ and $M_2$. One unit of type $A$ requires one minute of processing time on $M_1$ and two minutes of processing time on $M_2$, whereas one unit of type $B$ requires one minute of processing time on $\mathrm{M}_1$ and one minute on $\mathrm{M}_2$. Machines $\mathrm{M}_1$ and $\mathrm{M}_2$ are respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product should the firm produce a day in order to maximize the profit. Solve the problem graphically.
Answer
Let required number of product A and B be x and y respectively.
Since, profit on each product A and B are Rs. 5 and Rs. 3 respectively, so, profits on x product A and y product B are Rs. 5x and Rs. 3y respectively
Let Z be total profit so
$Z = 5x + 3y$
Since each unit of product A and B require one min. each on machine $M_1$
So, x unit of product A and y units of product B require x and y min. respectivley on machine $M_1$ but $M_1$ can work at most 5 x $60= 300$ min., so
$x +y ≤ 300$ (first constraint)
Since each unit of product A and B require 2 and one min. respectively on machine$ M_2.$
S0, x unit of product A and y units of product B require 2x and y min. respectivley on machine $M_2 $but $M_2$ can work at most 6 × $60 = 360$ min., so
$2x +y ≤ 360$ (second constraint)
Hence, mathematical formulation of LPP is find x and y which
maximize $Z = 5x + 3y$
Subject to constriants,
$x + y ≤ 300$
$= 2x + y ≤ 360$
$x, y ≥ 0 $ [Since production can not be less than zero]
Region $x + y ≤ 300:$
Line $x + y = 300$ meets axes at $A_1(300, 0), B_1(0, 300)$ respectively,
Region containing origin represents $x +y ≤ 300$ as (0, 0) satisfies $x + y = 300.$
Region $2x + y ≤ 360:$
Line $2x + y = 360$ meets axes at $A_2(180, 0), B_2(0, 360)$ respectively.
Region containing origin represents $2x + y ≤ 360$ as $(0, 0)$ satisfies $2x + y ≤ 360$
Region $x, y ≥ 0:$
It represent first quandrant
Shaded region $OA_2PB_2,$ represents feasible region.
Point P(60, 240) is obtained by solving
$x +y = 300$ and $2x + y = 360$

The value of $Z = 5x + 3y$ at
$O(0, 0) = 5(0) + 3(0) = 0$
$A_2(180, 0) = 5(180) + 3(0) = 900$
$P(60, 240) = 5(60) + 3(240) = 1020$
$B_1(0, 300) = 5(0) + 3(300) = 900$
Maximum $Z = 1020$ at $x = 60, y = 240$
Number of product $A = 60,$ product $ B = 240$
Maximum profit $= Rs. 1020.$
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Question 295 Marks
To maintain one's health, a person must fulfil certain minimum daily requirements for the following three nutrients: calcium, protein and calories. The diet consists of only items I and II whose prices and nutrient contents are shown below:
  Food I Food II Minimum daily requirement
Calcium 10 4 20
Protein 5 6 20
Calories 2 6 12
Price Rs. 0.60 per unit Rs. 1.00 per unit  
Find the combination of food items so that the cost may be minimum.
Answer
Let the person takes x units and y units of food I and II respectively that were taken in the diet.Since, per unit of food I costs Rs. 0.60 and that of food II costs Rs. 1.00.
Therefore, x lbs of food I costs Rs. 0.60 x and y lbs of food II costs Rs. 1.00y.
Total cost per day = Rs. (0.60x + 1.00y)
​Let Z denote the total cost per day
Then, Z = 0.60x + 1.00y
Since, each unit of food I contains 10 units of calcium.
Therefore, x units of food I contains 10x units of calcium.
Each unit of food II contains 4 units of calcium.
So, y units of food II contains 4y units of calcium.
Thus, x units of food I and y units of food II contains (10x + 4y) units of calcium.
But, the minimum requirement is 20 units of calcium.
$\therefore10\text{x}+4\text{y}\geq20$
Since, each unit of food I contains 5 units of protein.
Therefore, x units of food I contains 5x units of protein.
Each unit of food II contains 6 units of protein.
So, y units of food II contains 6y units of protein.
Thus, x units of food I and y units of food II contains (5x + 6y) units of protein.
But, the minimum requirement is 20 lbs of protein.
$\therefore5\text{x}+6\text{y}\geq20$
Since, each unit of food I contains 2 units of calories.
Therefore, x units of food I contains 2x units of calories.
Each unit of food II contains 6 units of calories.
So, y units of food II contains 6y units of calories.
Thus, x units of food I and y units of food II contains (2x + 6y) units of calories.
But, the minimum requirement is 12 lbs of calories.
$\therefore2\text{x}+6\text{y}\geq12$
Finally, the quantities of food I and food II are non negative values.
So, $\text{x},\text{y}\geq0$
Hence, the required LPP is as follow:
Min Z = 0.66x + 1.00y
Subject to
$10\text{x}+4\text{y}\geq20$
$5\text{x}+6\text{y}\geq20$
$2\text{x}+6\text{y}\geq12$
$\text{x},\text{y}\geq0$
First, we will convert the given inequations into equations, we obtain the following equations:
10x + 4y = 20, 5x +6y = 20, 2x + 6y = 12, x = 0 and y = 0
Region represented by 10x + 4y ≥ 20:
The line 10x + 4y = 20 meets the coordinate axes at A(2, 0) and B(0, 5) respectively.
By joining these points we obtain the line 10x + 4y = 20.
Clearly (0, 0) does not satisfies the inequation 10x + 4y ≥ 20.
So, the region in xy plane which does not contain the origin represents the solution set of the inequation 10x + 4y ≥ 20.
Region represented by $5\text{x}+6\text{y}\geq20$:
The line 5x + 6y = 20 meets the coordinate axes at C(4, 0) and $\text{D}\Big(0,\frac{10}{3}\Big)$ respectively.
By joining these points we obtain the line 5x + 6y = 20.
Clearly (0, 0) does not satisfies the inequation $5\text{x}+6\text{y}\geq20$.
So, the region in xy plane which does not contain the origin represents the solution set of the inequation $5\text{x}+6\text{y}\geq20$.
Region represented by 2x + 6y ≥ 12:
The line 2x + 6y =12 meets the coordinate axes at E(6, 0) and F(0, 2) respectively.
By joining these points we obtain the line 2x + 6y =12.
Clearly (0, 0) does not satisfies the inequation 2x + 6y ≥ 12.
So, the region which does not contains the origin represents the solution set of the inequation 2x + 6y ≥ 12.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 10x + 4y ≥ 20, 5x +6y ≥ 20, 2x + 6y ≥ 12, x ≥ 0, and y ≥ 0 are as follows.

The set of all feasible solutions of the above LPP is represented by the feasible region shaded in the graph.
The corner points of the feasible region are B(0, 5), $\text{G}\Big(1,\frac{5}{2}\Big),\text{H}\Big(\frac{8}{3},\frac{10}{9}\Big)$ and E(6, 0).
The value of the objective function at these points are given by the following table:
$\text{Points}$ $\text{Value of Z}$
$\text{B}$ $0.6(0)+5=5$
$\text{G}$ $0.6(1)+\frac{5}{2}=3.1$
$\text{H}$ $0.6\Big(\frac{8}{3}\Big)+\Big(\frac{10}{9}\Big)=1.6+1.1=2.7$
$\text{E}$ $0.6(6)+(0)=3.6$
We see that the minimum cost is 2.7 which is at $\Big(\frac{8}{3},\frac{10}{9}\Big)$.
Thus, at minimum cost, $\frac{8}{3}$ unit of food I and $\frac{10}{9}$ units of food II should be included in the diet.
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Question 305 Marks
A company sells two different products A and B. The two products are produced in a common production process and are sold in two different markets. The production process has a total capacity of 45000 man-hours. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The market has been surveyed and company officials feel that the maximum number of units of A that can be sold is 7000 and that of B is 10,000. If the profit is Rs. 60 per unit for the product A and Rs. 40 per unit for the product B, how many units of each product should be sold to maximize profit? Formulate the problem as LPP.
Answer
Product
Man hours
Maximum demand
Profit
A
5
7000
60
B
3
10000
40
Total capacity
45000
    
Let required production of product A be x units and production of product B be y units.

Given, profits on one unit of product A and B are Rs. 60 and Rs. 40 respectively, so profits on x units of product A and y units of product B are Rs. 60x and Rs. 40y.

Let Z be the total profit, so

Z = 60x + 40y

Given, production of one unit of product A and B require 5 hours and 3 hours respectively man hours, so x unit of product A and y units of product 8 require 5x hours and 3y hours of man hours respectively but total man hours available are 45000 hours, so

5x + 3y = 45000 (First constraint)

Given, dem and for product A is maximum 7000, so

$\times\leq7000$ (Second constraint)

Hence, mathematical formulation of LPP is,

Find x and y which

maximize Z = 60x + 40y

Subject to constraints,

$5\text{x}+3\text{y}\leq45000$

$\text{x}\leq7000$

$\text{y}\leq10000$

$\text{x},\text{y}\geq0$ [Since production can not be less than zero].
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Question 315 Marks
A furniture manufacturing company plans to make two products : chairs and tables. From its available resources which consists of 400 square feet to teak wood and 450 man hours. It is known that to make a chair requires 5 square feet of wood and 10 man-hours and yields a profit of Rs. 45, while each table uses 20 square feet of wood and 25 man-hours and yields a profit of Rs. 80. How many items of each product should be produced by the company so that the profit is maximum?
Answer
Let x units of chairs and y units of tables were produced

Therefore, $\text{x},\text{y}\geq0$

The given information can be tabulated as follows:
 
Wood (square feet)
Man hours
Chairs (x)
5
10
Tables (y)
20
25
Availability
400
450
Therefore, the constraints are

$5\text{x}+20\text{y}\leq400$

$10\text{x}+25\text{y}\leq450$

It is known that to make a chair requires 5 square feet of wood and 10 man-hours and yields a profit of Rs. 45, while each table uses 20 square feet of wood and 25 man-hours and yields a profit of Rs. 80.

Therefore, profit gained to make x chairs and y tables is Rs. 45x and Rs. 80y respectively Total profit = Z = 45x +80y which is to be maximised.

Thus, the mathematical formulation of the given linear programming problem is

Max 2 = 45x + 80y

Subject to

$5\text{x}+20\text{y}\leq400$

$10\text{x}+25\text{y}\leq450$

$\text{x},\text{y}\geq0$

First we will convert inequations into equations as follows:

5x + 20y = 400, 10x + 25y = 450, X = 0 and y = 0

Region represented by $5\text{x}+20\text{y}\leq400$:

The line 5x + 20y = 400 meets the coordinate axes at A(80, 0) and B(0, 20) respectively.

By joining these points we obtain the line 5x + 20y = 400.

Clearly (0, 0) satisfies the $5\text{x}+20\text{y}\leq400$.

So, the region which contains the origin represents the solution set of the inequation $5\text{x}+20\text{y}\leq400$.

Region represented by $10\text{x}+25\text{y}\leq450$:

The line 10x + 25y = 450 meets the coordinate axes at C(45, 0) and D(0, 18) respectively.

By joining these points we obtain the line 10x + 25y = 450.

Clearly (0, 0) satisfies the inequation $10\text{x}+25\text{y}\leq450$.

So the region which contains the origin represents the solution set of the inequation $10\text{x}+25\text{y}\leq450$.

Region represented by $\text{x}\geq0$ and $\text{y}\geq0$:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations $\text{x}\geq0$, and $\text{y}\geq0$.

The feasible region determined by the system of constraints $5\text{x}+20\text{y}\leq400,10\text{x}+25\text{y}\leq450,\text{x}\geq0,$ and $\text{y}\geq0$ are as follows.



The corner points are A(0, 18), B(45, 0)

The values of Z at these corner points are as follows.
Corner point
Z = 45x + 80y
A
1440
B
2025
The maximum value of Z is 2025 which is attained at B(45, 0).

Thus, the maximum profit is of Rs. 2025 obtained when 45 units of chairs and no units of tables are produced.
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Question 325 Marks
A company is making two products A and B. The cost of producing one unit of products A and B are Rs 60 and Rs 80 respectively. As per the agreement, the company has to supply at least 200 units of product B to its regular customers. One unit of product A requires one machine hour whereas product B has machine hours available abundantly within the company. Total machine hours available for product A are 400 hours. One unit of each product A and B requires one labour hour each and total of 500 labour hours are available. The company wants to minimize the cost of production by satisfying the given requirements. Formulate the problem as a LPP.
Answer
Let the company produces x units of product A and y units of product B.
Since, each unit of product A costs Rs. 60 and each unit of product B costs Rs. 80.Therefore, x units of product A and y units of product B will cost Rs. 60x and Rs 80y respectively.
Let Z denotes the total cost.
$\therefore$ Z = Rs. (60x + 80y)
Also, one unit of product A requires one machine hour.
The total machine hours available with the company for product A are 400 hours.
$\therefore\text{x}\leq400$
This is our first constraint
Also, one unit of product A and B require 1 labour hour each and there are a total of 500 labours hours.
Thus, $\text{x}+\text{y}\leq500$
​This is our second constraint.
Since, x and y are non negative integers, therefore $\text{x},\text{y}\geq\text{x},\text{y}\geq00$
Also, as per agreement, the company has to supply atleast 200 units of product B to its regular customers.
$\therefore\text{y}\geq200$
Hence, the required LPP is as follows:
Minimize Z = 60x + 80y
Subject to
$\text{x}\leq400$
$\text{x}+\text{y}\leq500$
$\text{y}\geq200$
$\text{x},\text{y}\geq0$
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Question 335 Marks
A diet is to contain at least 80 units of vitamin $A$ and 100 units of minerals. Two foods $F_1$ and $F_2$ are available. Food $F_1$ costs Rs 4 per unit and $F_2$ costs Rs 6 per unit one unit of food $F_1$ contains 3 units of vitamin $A$ and 4 units of minerals. One unit of food $F_2$ contains 6 units of vitamin $A$ and 3 units of minerals. Formulate this as a linear programming problem and find graphically the minimum cost for diet that consists of mixture of these foods and also meets the mineral nutritional requirements.
Answer
Let the dietician wishes to mix x units of food $F_1$ and y kg of $F_2.$
Clearly, $\text{x},\text{y}\geq0$
The given information can be tabulated as follows:
 
Vitamin A
Vitamin B
Food $F_1$
 3 4
Food $F_2$
 6 6 3
Minimum requirement 80 100
The constraints are
$3\text{x}+6\text{y}\geq80$
$4\text{x}+3\text{y}\geq100$
It is given that cost of food $F_1$ and $F_2$ is Rs. 4 and Rs. 6 per unit respectively.
Therefore, cost of x units of food $F_1$ and y units of food $F_2$ is Rs. 4x and Rs. 6y respectively.
Let Z denote the total cost
$\therefore Z = 4x + 6y$
Thus, the mathematical formulat​ion of the given linear programmimg problem is Minimize $Z = 4x+6y$
subject to
$3\text{x}+6\text{y}\geq80$
$4\text{x}+3\text{y}\geq100$
$\text{x},\text{y}\geq0$
First, we will convert the given inequations into equations, we obtain the following equations:
$3x + 6y = 80, 4x + 3y = 100, x = 0$ and $y = 0$
The line $3x + 6y = 80$ meets the coordinate axis at $\text{A}\Big(\frac{80}{3},0\Big)$ and $\text{B}\Big(0,\frac{40}{3}\Big)$.
Join these points to obtain the line $3x + 6y= 80.$
Clearly, $(0, 0)$ does not satisfies the inequation $3\text{x}+6\text{y}\geq80$.
So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
The line $4x + 3y = 100$ meets the coordinate axis at C(25, 0) and $\text{D}\Big(0,\frac{100}{3}\Big)$.
Join these points to obtain the line $4x + 3y = 100.$
Clearly, (0, 0) does not satisfies the inequation $4\text{x}+3\text{y}\geq100$.
So, the region in xyplane that does not contains the origin represents the solution set of the given equation.
Region represented by $x ≥ 0$ and $y ≥ 0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.
The feasible region determined by the system of constraints is

The corner points are $\text{D}\Big(0,\frac{100}{3}\Big),\text{E}\Big(24,\frac{4}{3}\Big)$ and $\text{A}\Big(\frac{80}{3},0\Big)$.
The values of Z at these corner points are as follows:
$\text{Corner point}$ $\text{Z}=4\text{x}+6\text{y}$
$\text{D}\Big(0,\frac{100}{3}\Big)$ $200$
$\text{E}\Big(24,\frac{4}{3}\Big)$ $104$
$\text{A}\Big(\frac{80}{3},0\Big)$ $\frac{320}{3}$
The minimum value of Z is Rs. 104 which is at $\text{E}\Big(24,\frac{4}{3}\Big)$.
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Question 345 Marks
A factory owner purchases two types of machines, A and B, for his factory. The requirements and limitations for the machines are as follows:
 
Area occupied by the
machine
Labour force for each
machine
Daliy outputin
units
Machines
1000 sp.m
12 mem
60
Machines
1200 sp.m
8 mem
40
He has an area of 7600 sq. m available and 72 skilled men who can operate the machines.
How many machines of each type should he buy to maximize the daily output?
Answer
Let required number of machine A and B are x and y respectively.
Since, production of each machine A and B are 60 and 40 units daily respectively.
So, productions by x number of machine A and y number of machine B are 60x and 40y respectively,
Let z denote total output daily, so,
$Z = 60x + 40y$
Since, each machine of type A and B require 1000 sq.m and 1200 sq.m area so, x machine of type A and y machine of type B require 100x and 1200y sq.m area but, Total area available for machine is 7600 sq.m. so,
$1000\text{x}+1200\text{y}\leq7600$
$5\text{x}+6\text{y}\leq38$ (first constraint)
Since, each machine of type A and B require 12 men and 8 men to work respectively so, x machine of type A and y machine of type B require 12x and By men to work respectively but, Total 72 men available for work so,
$12\text{x}+8\text{y}\leq72$
$3\text{x}+2\text{y}\leq18$ (second constraint)
Hence, mathematical formulation of LPP is,
Find x and y which maximize
$Z = 60x + 40y$
Subject to constraints,
$5\text{x}+6\text{y}\leq38$
$3\text{x}+2\text{y}\leq18$
$\text{x},\text{y}\geq0$ [Number of machines can not be less than zero]
Region $5\text{x}+6\text{y}\leq38$:
Line $5x + 6y = 38$ meets axes at $\text{A}_2\Big(\frac{38}{5},0\Big),\text{B}_1\Big(0,\frac{19}{3}\Big)$ respectively.
Region containing origin represents $5\text{x}+6\text{y}\leq38$ as origin satisfies $5\text{x}+6\text{y}\leq38$.
Region $3\text{x}+2\text{y}\leq18$:
line $3x + 2y = 18$ meets axes at $A_2(6, 0), B_2(0, 9)$ respectively.
Region containing origin represents $3x + 2y = 18$ as $(0, 0)$ satisfies $3\text{x}+2\text{y}\leq18$.
Region $\text{x},\text{y}\geq0$:
It represents first quadrant.

Shaded region $OA_2PB_1$, is the feasible region $P(4, 3)$ is obtained by solving $3x + 2y = 18$ and $5x+6y - 38$
The value of $Z = 60x + 40$y at
$\text{O}(0, 0) = 60(0) + 40(0) = 0$
$\text{A}_2(6, 0) = 60(6) + 40(0) = 360$
$\text{P}(4,3) =60(4)+ 40 (3) = 360$
$\text{B}_1\Big(0,\frac{19}{3}\Big)=60(0)+ 40\Big(\frac{19}{3}\Big)=\frac{760}{3}$
Therefore maximum $2 = 360$ at $x = 4, y = 3$ or $x = 6, y = 0$
Output is maximum when 4 machines of type A and 3 machine of type B or 6 machines of type A and no machine of type B.
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Question 355 Marks
A man owns a field of area 1000 sq.m. He wants to plant fruit trees in it. He has a sum of Rs. 1400 to purchase young trees. He has the choice of two types of trees. Type A requires 10 sq.m of ground per tree and costs Rs. 20 per tree and type B requires 20 sq.m of ground per tree and costs Rs. 25 per tree. When fully grown, type A produces an average of 20kg of fruit which can be sold at a profit of Rs. 2.00 per kg and type B produces an average of 40kg of fruit which can be sold at a profit of Rs. 1.50 per kg. How many of each type should be planted to achieve maximum profit when the trees are fully grown? What is the maximum profit?
Answer
Let the man planted x trees of type A and y trees of type B.

Number of trees cannot be negative.

Therefore, x, y ≥ 0

To plant tree of type A requires 10 sq. m and type B requires 20 sq. m of ground per tree.

And, it is given that a man owns a field of area 1000 sq. m.

Therefore,

10x + 20y ≤ 1000

Type A costs Rs 20 per tree and type B costs Rs. 25 per tree.

Therefore, x trees of type A and y trees of type B costs Rs. 20x and Rs. 25y respectively.

A man has a sum of Rs 1400 to purchase young trees.

20x + 25y ≤ 1400

Thus, the mathematical formulation of the given linear programmimg problem is

Max Z = 40x - 20x + 60y - 25y = 20x + 35y

Subject to

10x + 20y ≤ 1000, 20x + 25y ≤ 1400

The feasible region determined by the system of constraints is



The corner points are A(0, 50), B(20, 40), C(70, 0)

The values of Z at these corner points are as follows:
Corner point
Z = 20x + 35y
A
1750
B
1800
C
1400
The maximum value of Z is 1800 which is attained at B(20, 40)

Thus, the maximum profit is Rs. 1800 obtained when Rs. 20 were invested on type A and Rs. 40 were invested on type B.
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Question 365 Marks
Maximum Z = 30x + 20y Subject to $\text{x}+\text{y}\leq8$ $\text{x}+4\text{y}\geq12$ $5\text{x}+8\text{y}=20$$\text{x},\text{y}\geq0$
Answer
First, we will convert the given inequations into equations, we obtain the following equations: x + y = 8, x + 4y = 12, x = 0 and y = 0 5x + 8y = 20 is already an equation. Region represented by x + y ≤ 8: The line x + y = 8 meets the coordinate axes at A(8, 0) and B(0, 8) respectively. By joining these points. we obtain the line x + y = 8. Clearly (0, 0) satisfies the inequation x + y ≤ 8. So,the region in xy plane which contain the origin represents the solution set of the inequation x + y ≤ 8.Region represented by x + 4y ≥ 12:
The line x + 4y = 12 meets the coordinate axes at C(12, 0) and D(0, 3) respectively.
By joining these points we obtain the line x + 4y = 12.Clearly (0,0) satisfies the inequation x + 4y ≥ 12.
So,the region in xy plane which does not contain the origin represents the solution set of the inequation x + 4y ≥ 12.
The line 5x + 8y = 20 is the line that passes through E(4, 0) and $\text{F}\Big(0,\frac{5}{2}\Big)$.
Region represented by x ≥ 0 and y ≥ 0: Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0 and y ≥ 0. The feasible region determined by the system of constraints, x + y ≤ 8, x + 4y ≥ 12, 5x + 8y = 20, x ≥ 0 and y ≥ 0 are as follows.
The corner point of the feasible region are B(0, 8), D(0, 3), $\text{G}\Big(\frac{20}{3},\frac{4}{3}\Big)$. The value of Z at these corner points are as follows.
$\text{Corner point}$
$\text{Z}=30\text{x}+20\text{y}$
$\text{B}(0,8)$
$160$
$\text{D}(0,3)$
$60$
$\text{G}\Big(\frac{20}{3},\frac{4}{3}\Big)$
$266.66$
Therefore, the minimum value of Z is 60 at the point D(0, 3). Hence, x = 0 and y = 3 is the optimal solution of the given LPP. Thus, the optimal value of Z is 60.
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Question 375 Marks
A manufacturer produces two types of steel trunks. He has two machines A and B. For completing, the first types of the trunk requires 3 hours on machine A and 3 hours on machine B, whereas the second type of the trunk requires 3 hours on machine A and 2 hours on machine B. Machines A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of Rs. 30 and Rs. 25 per trunk of the first type and the second type respectively. How many trunks of each type must he make each day to make maximum profit?
Answer
Let x trunks of first type and y trunks of second type were manufactured.
Number of trunks cannot be negative.
Therefore,
$x, y ≥ 0$
According to question, the given information can be tabulated as
 
Machine A (hrs)
Machine B (hrs)
First type (x)
3
3
Second type (y)
3
2
Availability
18
15
Therefore, the constraints are
$3\text{x}+3\text{y}\leq18$
$3\text{x}+2\text{y}\leq15$
He earns a profit of Rs. 30 and Rs. 25 per trunk of the first type and the second type respectively.
Therefore, profit gained by him from x trunks of first type and y trunks of second type is Rs. 30x and Rs. 25y respectively.
Total profit = Z = 30x + 25y which is to be maximised
Thus, the mathematical formulat​ion of the given linear programmimg problem is
Max $Z = 30x + 25y$
Subject to
$3\text{x}+3\text{y}\leq18$
$3\text{x}+2\text{y}\leq15$
$\text{x},\text{y}\geq0$
First we will convert inequations into equations as follows:
$3x + 3y = 18, 3x + 2y = 15, x = 0$ and $y = 0$
Region represented by $3x + 3y ≤ 18:$
The line 3x + 3y = 18 meets the coordinate axes at $A_1(6, 0)$ and $B_1(0, 6)$ respectively.
By joining these points we obtain the line $3x + 3y = 18$.
Clearly (0, 0) satisfies the $3x + 3y = 18.$
So, the region which contains the origin represents the solution set of the inequation $3x + 3y ≤ 18.$
Region represented by $3x + 2y ≤ 15:$
The line $3x + 2y = 15$ meets the coordinate axes at $C_1(5, 0)$ and $\text{D}_1\Big(0,\frac{15}{2}\Big)$ respectively.
By joining these points we obtain the line $3x + 2y = 15.$
Clearly (0, 0) satisfies the inequation $3x + 2y ≤ 15.$
So, the region which contains the origin represents the solution set of the inequation $3x + 2y ≤ 15.$
Region represented by $x ≥ 0$ and $y ≥ 0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $x ≥ 0,$ and $y ≥ 0.$
The feasible region determined by the system of constraints $3x + 3y ≤ 18, 3x + 2y ≤ 15, x ≥ 0$ and $y ≥ 0$ are as follows.

The corner points are $O(0, 0), B_1(0, 6), E_1(3, 3)$ and $C_1(5, 0).$
The values of Z at these corner points are as follows.
Corner point
$Z = 30x + 25y$
$O$
$0$
$B_1$
$15$
$E_1$
$165$
$C_1$ $150$
The maximum value of Z is 165 which is attained at $E_1(3, 3).$
Thus, the maximum profit is Rs. 165 obtained when 3 units of each type of trunk is manufactured.
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Question 385 Marks
A company sells two different products, A and B. The two products are produced in a common production process, which has a total capacity of 500 man-hours. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The market has been surveyed and company officials feel that the maximum number of unit of A that can be sold is 70 and that for B is 125. If the profit is Rs. 20 per unit for the product A and Rs. 15 per unit for the product B, how many units of each product should be sold to maximize profit?
Answer
Let x units of product A and y units of product B were manufactured.
Clearly, $x ≥ 0, y ≥ 0$
It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B.
The two products are produced in a common production process, which has a total capacity of 500 man-hours.
$5x + 3y ≤ 500$
The maximum number of unit of A that can be sold is 70 and that for B is 125.
$X ≤ 70$
$y ≤ 125$
If the profit is Rs. 20 per unit for the product A and Rs. 15 per unit for the product B.
Therefore, profit x units of product A and y units of product B is Rs. 20x and Rs. 15y respectively.
Total profit $= Z = 20x + 15y$
The mathematical formulation of the given problem is
Max $Z = 20x + 15y$
Subject to
$5x + 3y ≤ 500$
$x ≤ 70$
$y ≤ 125$
$x ≥ 0$
$y ≥ 0$
First we will convert inequations into equations as follows:
$5x + 3y = 500, x = 70, y = 125, x = 0$ and $y = 0$
Region represented by $5x + 3y ≤ 500:$
The line $5x + 3y = 500$ meets the coordinate axes at $A_1(100, 0)$ and $B_1(0,5003)$ respectively.
By joining these points we obtain the line $5x + 3y = 500.$
Clearly $(0, 0)$ satisfies the $5x + 3y = 500.$
So, the region which contains the origin represents the solution set of the inequation $5x + 3y ≤ 500.$
Region represented by $x ≤ 70:$
The line $x = 70$ is the line passes through $C_1(70, 0)$ and is parallel to Y axis.
The region to the left of the line $x = 70$ will satisfy the inequation $x ≤ 70.$
Region represented by $y ≤ 125$:
The line y = 125 is the line passes through $D_1(0, 125)$ and is parallel to X axis.
The region below the the line $y = 125$ will satisfy the inequation $y ≤ 125.$
Region represented by $x ≥ 0$ and $y ≥ 0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $x ≥ 0$, and $y ≥ 0.$
The feasible region determined by the system of constraints $5x + 3y ≤ 500, x ≤ 70, y ≤ 125, x ≥ 0$ and $y ≥ 0$ are as follows.

The corner points are $O(0, 0), D_1(0, 125), E_1(25, 125), F_1(70, 50)$ and $C_1(70, 0).$
$$The values of Z at the corner points are
Corner points
Z = 20x + 15y
$O$
0
$D_1$
1875
$E_1$
2375
$F_1$
2150
$C_1$
1400
The maximum value of Z is 2375 which is at $E_1(25, 125).$
Thus, maximum profit is Rs. 2375, 25 units of A and 125 units of B should be manufactured.
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Question 395 Marks
Kellogg is a new cereal formed of a mixture of bran and rice that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligrams of iron per kilogram, and that rice contains 100 grams of protein and 30 milligrams of iron per kilogram, find the minimum cost of producing this new cereal if bran costs Rs. 5 per kg and rice costs Rs 4 per kg.
Answer
Let required quantity of bran and rice be x kg and y kg.

Given, costs of one kg of bran and rice are Rs. 5 and Rs. 4 per kg,

So, costs of X unit of bran and Y kg of rice are 5x and Rs 4y respectively,

Let total cost of bran and rice be Z, so,

Z = 5x + 4y

Since one kg of bran and rice contain 80 and 100 mg of protien, so, x kg of bran and y kg of rice contain 80x and 100y grms of protien respectively, but minimum requirement of protien for kelloggs is 88 gms, so

$80\text{x}+100\text{y}\geq88$

$20\text{x}+25\text{y}\geq22$ (first constraint)

Since one kg of bran and rice contain 40 mg and 30 mg of iron, so, x kg of bran and y kg of rice contain 40x and 30y mg of iron respectively, but minimum requirement of iron is 36 mg for kelloggs, so $40\text{x}+30\text{y}\geq36$

$40\text{x}+30\text{y}\geq36$ (second constraint)

Hence, mathematical formulation of LPP is,

Find x and y which minimize

Z = 5x + 4y

subject to constraints,

$20\text{x}+25\text{y}\geq22$

$40\text{x}+30\text{y}\geq36$

$\text{x},\text{y}\geq0$ [Since quantity of bran and rice can not be less than zero]

Region $20\text{x}+25\text{y}\geq22$: line 20x + 25y = 22 meets axes at $\text{A}_1\Big(\frac{11}{10},0\Big),\text{B}_1\Big(0,\frac{22}{25}\Big)$ espectively.

Region not containing origin represents $20\text{x}+25\text{y}\geq22$ as (0, 0) does not satisfy $20\text{x}+25\text{y}\geq22$.

Region $40\text{x}+30\text{y}\geq36$ line 40x + 30y = 36 meets axes at $\text{A}_1\Big(\frac{9}{10},0\Big),\text{B}_1\Big(0,\frac{6}{5}\Big)$

Region not containing origin represents $40\text{x}+30\text{y}\geq36$ as (0, 0) does not satisfy $40\text{x}+30\text{y}\geq36$.



The value of Z = 5x + 4y at

$\text{A}_1\Big(\frac{11}{10},0\Big)=5\Big(\frac{11}{10}\Big)+4(0)=5.5$

$\text{P}\Big(\frac{3}{5},\frac{2}{5}\Big)=5\Big(\frac{3}{5}\Big)+4\Big(\frac{6}{5}\Big)=4.6$

$\text{B}_2\Big(0,\frac{6}{5}\Big)=5(0)+4\Big(\frac{6}{5}\Big)=4.8$

Smallest value of Z is 4.6.

Now open half plane 5x + 4y < 4.6 has no point in common with feasible region so, smallest value z is the minimum value.

Hence,

Minimum cost of mixture = Rs 4.6
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Question 405 Marks
A firm manufacturing two types of electric items, A and B, can make a profit of Rs. 20 per unit of A and Rs. 30 per unit of B. Each unit of A requires 3 motors and 4 transformers and each unit of B requires 2 motors and 4 transformers. The total supply of these per month is restricted to 210 motors and 300 transformers. Type B is an export model requiring a voltage stabilizer which has a supply restricted to 65 units per month. Formulate the linear programing problem for maximum profit and solve it graphically.
Answer
Let x units of item A and y units of item B were manufactured.
Number of items cannot be negative.
Therefore, $\text{x},\text{y}\geq0$
The given information can be tabulated as follows:
Product
Motors
Transformers
A(x)
3
4
B(y)
2
4
Availability
210
300
Further, it is given that type B is an export model, whose supply is restricted to 65 units per month.
Therefore, the constraints are
$3\text{x}+2\text{y}\leq210$
$4\text{x}+4\text{y}\leq300$
$\text{y}\leq65$
A and B can make a profit of Rs. 20 per unit of A and Rs. 30 per unit of B.
Therefore, profit gained from x units of item A and y units of item B is Rs. 20x and Rs. 30y respectively.
Total profit $= Z = 20x + 30y$ which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max $Z = 20x + 30y$
Subject to
$3\text{x}+2\text{y}\leq210$
$4\text{x}+4\text{y}\leq300$
$\text{y}\leq65$
$\text{x},\text{y}\geq0$
First we will convert inequations into equations as follows:
$3x + 2y = 210, 4x + 4y = 300, y = 65, x = 0$ and $y = 0$
Region represented by $3\text{x}+2\text{y}\leq210$:
The line $3x + 2y = 210$ meets the coordinate axes at $A_1(70, 0)$ and $B_1(0, 105)$ respectively.
By joining these points we obtain the line $3x + 2y = 210.$
Clearly (0, 0) satisfies the $3x + 2y = 210.$
So, the region which contains the origin represents the solution set of the inequation $3\text{x}+2\text{y}\leq210$.
Region represented by $4\text{x}+4\text{y}\leq300$:
The line $4x + 4y = 300$ meets the coordinate axes at $C_1(75,0)$ and $D_1(0,75)$ respectively.
By joining these points we obtain the line $4x + 4y = 300.$
Clearly $(0, 0)$ satisfies the inequation $4\text{x}+4\text{y}\leq300$.
So,the region which contains the origin represents the solution set of the inequation $4\text{x}+4\text{y}\leq300$.
$y = 65$ is the line passing through the point $E_1(0, 65) $ and is parallel to X axis.
Region represented by $\text{x}\geq0$ and $\text{y}\geq0$:
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $\text{x}\geq0$, and $\text{y}\geq0$.
The feasible region determined by the system of constraints $3\text{x}+2\text{y}\leq210,4\text{x}+4\text{y}\leq300,\text{x}\geq0,$ and $\text{y}\geq0$ are as follows:

The corner points are $O(0, 0), E_1(0, 65), G_1(10, 65), F_1(60, 15)$ and $A_1(70, 0).$
The values of Z at these corner points are as follows.
Corner point
Z = 20x + 30y
$O$
$0$
$E_1$
$1950$
$G_1$
$2150$
$F_1$
$1650$
$A_1$
$1400$
The maximum value of Z is 2150 which is attained at $G(10, 65).$
Thus, the maximum profit is Rs. 2150 obtained when 10 units of item A and 65 units of item Bwere manufactured.
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Question 415 Marks
A factory uses three different resources for the manufacture of two different products, 20 units of the resources A, 12 units of B and 16 units of C being available. 1 unit of the first product requires 2, 2 and 4 units of the respective resources and 1 unit of the second product requires 4, 2 and 0 units of respective resources. It is known that the first product gives a profit of 2 monetary units per unit and the second 3. Formulate the linear programming problem. How many units of each product should be manufactured for maximizing the profit? Solve it graphically.
Answer
Let x units of first product and y units of second product be manufactured.
Therefore, x, y ≥ 0
The given information can be tabulated as follows:
Product
 Resource A Resource B Resource C
First(x)
 2 2 4
Second (y)
 4 2 0
Availability
20
 12 16
Therefore, the constraints are
$2\text{x}+4\text{y}\leq20$
$2\text{x}+2\text{y}\leq12$
$4\text{x}+0\text{y}\leq16$
$4\text{x}\leq16$
It is known that the first product gives a profit of 2 monetary units per unit and the second 3.
Therefore, profit gained from x units of first product and y units of second product is 2x monetary units and 4y monetary units respectively.
Total profit $= Z = 2x + 3$y which is to be maximised
Thus, the mathematical formulat​ion of the given linear programmimg problem is
Max $Z = 2x + 3y$
Subject to
$2\text{x}+4\text{y}\leq20$
$2\text{x}+2\text{y}\leq12$
$4\text{x}+0\text{y}\leq16$
$4\text{x}\leq16$
$\text{x},\text{y}\geq0$
First we will convert inequations into equations as follows:
$2x + 4y = 20, 2x + 2y = 12, 4x = 16, x = 0$ and $y = 0$
Region represented by $2x + 4y ≤ 20:$
The line $2x + 4y = 20$ meets the coordinate axes at $A_1(10, 0)$ and $B_1(0, 5)$ respectively.
By joining these points we obtain the line $2x + 4y = 20.$
Clearly (0, 0) satisfies the $3x + 2y = 210.$
So, the region which contains the origin represents the solution set of the inequation $2x + 4y ≤ 20.$
Region represented by $2x + 2y ≤ 12:$
The line $2x + 2y = 16$ meets the coordinate axes at $C_1(6, 0)$ and $D_1(0, 6)$ respectively.
By joining these points we obtain the line $2x + 2y = 12.$
Clearly (0, 0) satisfies the inequation $2x + 2y ≤ 12.$
So, the region which contains the origin represents the solution set of the inequation $2x + 2y ≤ 12.$
Region represented by $4x ≤ 16:$
The line $4x =16$ or $x = 4$ is the line passing through the point $E_1(4, 0)$ and is parallel to Y axis.
The region to the left of the line $x = 4$ would satisfy the inequation $4x ≤ 16.$
Region represented by $x ≥ 0$ and $y ≥ 0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $x ≥ 0$, and $y ≥ 0$.
The feasible region determined by the system of constraints $2x + 4y ≤ 20, 2x + 2y ≤ 12, 4x ≤ 16, x ≥ 0$ and $y ≥ 0 $ are as follows.

The corner points are $O, 0), B_1(0, 5), G_1(2, 4), F_1(4, 2)$ and $E_1(4, 0).(0$
The values of Z at these corner points are as follows.
Corner point
$Z = 2x + 3y$
$O$
$0$
${B_1}$
$15$
$G_1$
 $16$
$F_1$
$14$
$E_1$ $8$
The maximum value of Z is 16 which is attained at $G_1(2, 4)$
Thus, the maximum profit is 16 monetary units obtained when 2 units of first product and 4 units of second product were manufactured.
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Question 425 Marks
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:
Food
Vitamin A
Vitamin B
Vitamin C
X
1
2
3
Y
2
2
1
One kg of food X costs Rs. 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet?
Answer
Let x be the amount of food X and Y be the amount of food Y that is to be mixed which will produce the required diet.
Then the mathematical modal of the LPP is as follows:
Minimise $Z = 16x + 20y$
Subject to
$2\text{x}+2\text{y}\geq12$
$3\text{x}+\text{y}\geq8$
$\text{x}\geq0,\text{y}\geq0$
To solve the LPP we draw the lines,
$x + 2y = 10$
$2x + 2y = 12$
$3x + y = 8$
The feasible region of the LPP is shaded in graph.

The coordinates of the vertices (corner points) of the feasible region ABCD are $A(10, 0), B(2, 4), C(1, 5)$ and $D(0, 8$).
The value of the objective function at these points are given in the following table.
Point $(x_1, x_2)$
Value of objective function $Z = 16x + 20y$
$A(10, 0)$ $Z = 160$
$B(2, 4)$
$Z = 112$
$C(1, 5)$ $Z = 116$
$D(0, 8)$ $Z = 160$
2kg of food X and 4kg of food Y will be required to minimize the cost of the diet.
The least cost of the mixture is Rs. 112.
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Question 435 Marks
An airline agrees to charter planes for a group. The group needs at least 160 first class seats and at least 300 tourist class seats. The airline must use at least two of its model 314 planes which have 20 first class and 30 tourist class seats. The airline will also use some of its model 535 planes which have 20 first class seats and 60 tourist class seats. Each flight of a model 314 plane costs the company Rs 100,000 and each flight of a model 535 plane costs Rs 150,000. How many of each type of plane should be used to minimize the flight cost? Formulate this as a LPP.
Answer
Let x number of model 314 planes and y number of model 535 planes were used.

It is given that cost of one model 314 plane is Rs 100000 and cost of one model 535 plane is Rs 150000.

Therefore, cost of x model 314 plane is Rs 100000x and cost of y model 535 plane is Rs 150000y.

Total cost price = 100000x +150000 y

Let Z denote the total cost

Then, Z = 100000x +150000y

Also,

Each model 314 planes have 20 first class and 30 tourist class seats and each model 535 planes has 20 first class and 60 tourist class seats.

The group needs 160 first class seats and 300 tourist class seats.

$\therefore20\text{x}+20\text{y}\geq160,30\text{x}+60\text{y}\geq300$

Number of planes cannot be negative.

Therefore, $\text{x},\text{y}\geq0$

Hence, the required LPP is as follows:

Min Z = 100000x+150000y

Subject to

$\therefore20\text{x}+20\text{y}\geq160$

$30\text{x}+60\text{y}\geq300$

$\text{x}\geq0,\text{y}\geq0$
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Question 445 Marks
A firm makes items A and B and the total number of items it can make in a day is 24. It takes one hour to make an item of A and half an hour to make an item of B. The maximum time available per day is 16 hours. The profit on an item of A is Rs. 300 and on one item of B is Rs. 160. How many items of each type should be produced to maximize the profit? Solve the problem graphically.
Answer
Let the number of item A produced be x, and the number of item B produced be y. Since the total number of items are at most 24. $\text{x} + \text{y} \leq 24 \ \dots(1)$ Item A takes 1 hour to manufacture and item B takes half an hour to manufacture. x item takes x hour to manufacture and y items take $\frac{\text{y}}{2}$ hour to a manufacture. and maximum time available is 16 hours. $\therefore \text{x} + \frac{\text{y}}{2}\leq16 \dots (2)$ The profit on one unit of A is Rs. 300 and the profit on one unit of B is Rs. 160 Since we want to maximize profit. Let the profit be z. Max z = 300x + 160y .....(3)
The shaded region will satisfy the equation (1) and (2), their intersection point is E(8, 16). Vertices of OAED are O(0, 0), A(0, 24), E(8, 16) and D(16, 0). At A Z = 160 × 24 = 3840 At E Z = 300 × 8 + 160 × 16 = 2400 + 2560 = 4960 At D Z = 300 × 16 + 160 × 0 = 4800 Therefore the maximum value is at point E. i.e. x = 8 and y = 16 Thus they should produce 8 items of type A and 16 items of type B.
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Question 455 Marks
Maximum Z = 2x + 4y Subject to$\text{x}+\text{y}\geq8$
$\text{x}+4\text{y}\geq12$
$\text{x}\geq3,\text{y}\geq2$
Answer
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 8, x + 4y = 12, x = 3, y = 2
Region represented by x + y ≥ 8:
The line x + y = 8 meets the coordinate axes at A(8, 0) and B(0, 8) respectively. By joining these points we obtain the line x + y = 8.
Clearly (0, 0) does not satisfies the inequation x + y ≥ 8.
So, the region in xy plane which does not contain the origin represents the solution set of the inequation x + y ≥ 8.
Region represented by x + 4y ≥ 12:
The line x + 4y = 12 meets the coordinate axes at C(12, 0) and D(0, 3) respectively. By joining these points we obtain the line x + 4y = 12.
Clearly (0, 0) satisfies the inequation x + 4y ≥ 12.
So, the region in xy plane which contain the origin represents the solution set of the inequation x + 4y ≥ 12.
The line x = 3 is the line that passes through the point (3, 0) and is parallel to Y axis. x ≥ 3 is the region to the right of the line x = 3.

The line y = 2 is the line that passes through the point (0, 12) and is parallel to X axis. y ≥ 2 is the region above the line y = 2.

The corner points of the feasible region are E(3, 5) and F(6, 2).
The values of Z at these corner points are as follows.
Corner point
Z = 2x + 4y
E(3, 5)
2 × 3 + 4 × 5 = 26
F(6, 2)
2 × 6 + 4 × 2 = 20
Therefore, the minimum value of Z is 20 at the point F(6, 2).
Hence, x = 6 and y = 2 is the optimal solution of the given LPP.
Thus, the optimal value of Z is 20.
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Question 465 Marks
Show the solution zone of the following inequalities on a graph paper:
$5\text{x}+\text{y}\geq10$
$\text{x}+\text{y}\geq6$
$\text{x}+4\text{y}\geq12$
$\text{x}\geq,\text{y}\geq0$
Answer
Converting the given inequations into equations
5x + y = 10, x + y = 6, x + 4y = 12, x = y = 0

Region represented by $5\text{x}+\text{y}\geq10$:
Line $5x + y - 10$ meets coordinate axes at $A_1(2, 0)$ and $B_1(0, 10).$
Clearly, $(0, 0)$ does not satisfy $5\text{x}+\text{y}\geq10$, so region not containing origin represents $5\text{x}+\text{y}\geq10$ in xy -plane.
Region represented by $\text{x}+\text{y}\geq6$:
Line $x + y = 6$ meets coordinate axes at $A_2(6, 0)$ and $B_2(0, 6).$
Clearly, $(0, 0)$ does not satisfy $\text{x}+\text{y}\geq6$,
so region not containing origin represents $\text{x}+\text{y}\geq6$ in xy -plane.
Region represented by $\text{x}+4\text{y}\geq12$:
Line $x + 4y = 12$ meets coordinate axes at $A_3(12, 0)$ and $B_3(0, 3).$
Clearly, (0, 0) does not satisfy $\text{x}+4\text{y}\geq12$, so, region not containing origin $\text{x}+4\text{y}\geq12$ in xy - plane.
Region represented by $\text{x}\geq,\text{y}\geq0$:
It represents first quadrant in xy-plane.
The unbounded shaded region with corner points $A_3(12, 0), P(4, 2), Q(1, 5), B_1(0, 10)$ represents feasible region.
Point P is obtained by solving $x + 4y = 12$ and $x + y = 6, $ Q by solving $x + y = 6$ and $5x + y = 10.$
The value of $Z = 3x + 2y $at
$A_3(12, 0) = 3(12) + 2(0) = 36$
$P(4, 2) = 3(4) + 2(2) = 16$
$Q(1, 5) = 3(1) + 2(5) = 13$
$B(0, 10) = 3(0) + 2(10) = 20$
Smallest value of $Z = 13,$
Now open half plane $3\text{x}+2\text{y}\leq13$ has no point in common with feasible region, so, smallest value is the minimum value of Z,
Hence,
Minimum $z = 13$ at $x = 1, y = 5$
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Question 475 Marks
A company manufactures two articles A and B. There are two departments through which these articles are processed: (i) assembly and (ii) finishing departments. The maximum capacity of the first department is 60 hours a week and that of other department is 48 hours per week. The product of each unit of article A requires 4 hours in assembly and 2 hours in finishing and that of each unit of B requires 2 hours in assembly and 4 hours in finishing. If the profit is Rs. 6 for each unit of A and Rs 8 for each unit of B, find the number of units of A and B to be produced per week in order to have maximum profit.
Answer
Let x units and y units of articles A and B are produced respectively.
Number of articles cannot be negative.
Therefore, $x, y ≥ 0$
The product of each unit of article A requires 4 hours in assembly and that of article B requires 2 hours in assembly and the maximum capacity of the assembly department is 60 hours a week
$4x + 2y ≤ 60$
The product of each unit of article A requires 2 hours in finishing and that of article B requires 4 hours in assembly and the maximum capacity of the finishing department is 48 hours a week.
$2x + 4y ≤ 48$
If the profit is Rs. 6 for each unit of A and Rs. 8 for each unit of B.
Therefore, profit gained from​ x units and y units of articles A and B respectively is Rs. 6x and Rs. 8y respectively.
Total revenue $= Z = 6x + 8y$ which is to be maximised.
Thus, the mathematical formulat​ion of the given linear programmimg problem is
Max $Z = 6x + 8y$
Subject to
$2x + 4y ≤ 48$
$4x + 2y ≤ 60$
$x, y ≥ 0$
First we will convert inequations into equations as follows:
$2x + 4y = 48, 4x + 2y = 60, x = 0$ and $y = 0$
Region represented by $2x + 4y ≤ 48:$
The line $2x + 4y = 48$ meets the coordinate axes at $A_1(24, 0)$ and $B_1(0, 12)$ respectively.
By joining these points we obtain the line $2x + 4y = 48.$
Clearly (0, 0) satisfies the $2x + 4y = 48.$
So, the region which contains the origin represents the solution set of the inequation $2x + 4y ≤ 48.$
Region represented by $4x + 2y ≤ 60:$
The line $4x + 2y = 60$ meets the coordinate axes at $C_1(15, 0)$ and $D_1(0, 30)$ respectively.
By joining these points we obtain the line $4x + 2y = 60.$
Clearly (0, 0) satisfies the inequation $4x + 2y ≤ 60.$
So, the region which contains the origin represents the solution set of the inequation $4x + 2y ≤ 60.$
Region represented by $x ≥ 0$ and $y ≥ 0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $x ≥ 0$, and $y ≥ 0.$
The feasible region determined by the system of constraints $2x + 4y ≤ 48, 4x + 2y ≤ 60, x ≥ 0$ and $y ≥ 0$ are as follows.

The corner points are $O(0, 0), B_1(0, 12), E_1(12, 6)$ and $C_1(15, 0).$
The values of Z at these corner points are as follows.
Corner points
$Z = 6x + 8y$
$O$
$0$
$B_1$
 $96$
$E_1$
$120$
$C_1$
$90$
The maximum value of Z is 120 which is attained at $E_1(12, 6).$
Thus, the maximum profit is Rs. 120 obtained when 12 units of article A and 6 units of article B were manufactured.
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Question 485 Marks
A company produces two types of goods, A and B, that require gold and silver. Each unit of type A requires 3gm of silver and 1 gm of gold while that of type B requires 1 gm of silver and 2gm of gold. The company can produce 9gm of silver and 8gm of gold. If each unit of type A brings a profit of Rs. 40 and that of type B Rs. 50, find the number of units of each type that the company should produce to maximize the profit. What is the maximum profit?
Answer
Let number of goods A and B are x and y respectively.
Since, profits on each A and B are Rs. 40 and Rs. 50 respectively.
So, profits on x of type A and y of type B are 40x and 50y respectively, Let Z be total profit on A and B, so,
$Z = 40x + 50y$
Since, each A and B require 3gm and 1gm of silver respectively.
So, x of type A and y type B require 3x and y gm silver respectively but, Total silver available is 9 gm. so,
$3\text{x}+\text{y}\leq9$ (first constraint)
Since, each A and B require 1gm and 2gm of gold respectively.
So, x of type A and y type B require x and 2y gm of gold respectively but, Total gold available is 8 gm, so,
$\text{x}+2\text{y}\leq8$ (second constraint)
Hence, mathematical formulation of LPP is,
Find x and y which maximize
$Z = 40x + 50y$
Subject to constraints,
$3\text{x}+\text{y}\leq9$
$\text{x}+2\text{y}\leq8$
$\text{x},\text{y}\geq0$ [Since production of A and B can not be less than zero]
Region $3\text{x}+\text{y}\leq9$:
Line $3x +y = 9$ meets axes at $A(3, 0), B(0, 9)$ respectively.
Region containing origin represents $3\text{x}+\text{y}\leq9$ as $(0, 0)$ satisfies $3\text{x}+\text{y}\leq9$.
Region $\text{x}+2\text{y}\leq8$:
Line $x +2y = 8$ meets axes at $A_2(8, 0), B_2(0, 4)$ respectively.
Region containing origin represents $\text{x}+2\text{y}\leq8$ as $(0, 0)$ satisfies $\text{x}+2\text{y}\leq8$.
Region $\text{x},\text{y}\geq0$:
It represents first quadrant.

Shaded region $OA_2PB_2 $ is the feasible region.
Point P(2, 3) is obtained by solving $3x + y - 9$ and $x + 2y = 8$
The value of $Z = 40x + 50y$ at
$O(0, 0) = 40(0) + 50(0) = 0$
$A_1(3, 0) = 40(3) + 50(0) = 120$
$P(2, 3) = 40(2) + 50(3) = 230$
$B_2(0, 4) = 40(0) + 50(4) = 200$
Therefore maximum $Z = 230$ at $x = 2, y = 3$
Hence,
Maximum profit = Rs. $230$ number of goods of type $A = 2$, type $B = 3$
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Question 495 Marks
A small manufacturer has employed 5 skilled men and 10 semi-skilled men and makes an article in two qualities deluxe model and an ordinary model. The making of a deluxe model requires 2 hrs. work by a skilled man and 2 hrs. work by a semi-skilled man. The ordinary model requires 1 hr by a skilled man and 3 hrs. by a semi-skilled man. By union rules no man may work more than 8 hrs per day. The manufacturers clear profit on deluxe model is Rs. 15 and on an ordinary model is Rs. 10. How many of each type should be made in order to maximize his total daily profit.
Answer
Let x articles of deluxe model and y articles of an ordinary model be made.

Number of articles cannot be negative.

Therefore, $\text{x},\text{y}\geq0$

According to the question, the making of a deluxe model requires 2 hrs.

Work by a skilled man and the ordinary model requires 1 hr by a skilled man

$2\text{x}+\text{y}\leq40$

The making of a deluxe model requires 2 hrs.

Work by a semi-skilled man ordinary model requires 3 hrs.

Work by a semi-skilled man. $2\text{x}+3\text{y}\leq80$

Total profit = Z = 152 + 10y which is to be maximised

Thus, the mathematical formulation of the given linear programmimg problem is

Max Z = 15x + 10y

Subject to

$2\text{x}+\text{y}\leq40$

$2\text{x}+3\text{y}\leq80$

$\text{x}\geq0$

$\text{y}\geq0$

The feasible region determined by the system of constraints is:



The corner points are $\text{A}\Big(0,\frac{800}{3}\Big)$, B(10, 20), C(20, 0).

The values of Z at these corner points are as follows.
Corner point
Z = 15x + 10y
A
$\frac{800}{3}$
B
350
C
300
The maximum value of Z is 300 which is attained at C(20, 0).

Thus, the maximum profit is Rs. 300 obtained when 10 units of deluxe modal and 20 unit of ordinary model is produced.
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Question 505 Marks
Show the solution zone of the following inequalities on a graph paper:
$\text{x}+\text{y}\leq50$
$3\text{x}+\text{y}\geq90$
$\text{x},\text{y}\geq0$
Answer
We have to maximize Z = 60x + 15y.

First, we will convert the given inequations into equations, we obtain the following equations:

x + y = 50, 3x + y = 90, x = 0 and y = 0

Region represented by $\text{x}+\text{y}\leq50$.

The line x + y = 50 meets the coordinate axes at A(50, 0) and B(0, 50) respectively.

By joining these points we obtain the line 3x + 5y = 15.

Clearly (0, 0) satisfies the inequation $\text{x}+\text{y}\leq50$.

So, the region containing the origin represents the solution set of the inequation $\text{x}+\text{y}\leq50$.

Region represented by $3\text{x}+\text{y}\geq90$.

The line 3x + y = 90 meets the coordinate axes at C(30, 0) and D(0, 90) respectively.

By joining these points we obtain the line $3\text{x}+\text{y}\geq90$.

Clearly (0, 0) satisfies the inequation $3\text{x}+\text{y}\geq90$.

So, the region containing the origin represents the solution set of the inequation $3\text{x}+\text{y}\geq90$.

Region represented by $\text{x}\geq0$ and $\text{y}\geq0$:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations $\text{x}\geq0$ and $\text{y}\geq0$:

The feasible region determined by the system of constraints,

$\text{x}+\text{y}\leq50,3\text{x}+\text{y}\geq90,\text{x}\geq0$ and $\text{y}\geq0$, are as follows.



The corner points of the feasible region are O(0, 0), C(30, 0), E(20, 30 ) and B(0, 50).

The values of Z at these corner points are as follows.
Corner point Z = 60x + 15y
O(0, 0) 60 × 0 + 15 × 0 = 0
C(30, 0) 60 × 30 + 15 × 0 = 1800
E(20, 30) 60 × 20 + 15 × 30 = 1650
B(0, 50) 60 × 0 + 15 × 50 = 50
Therefore, the maximum value of Z is 1800 at the point (30, 0).

Hence, x = 30 and y = 0 is the optimal solution of the given LPP.

Thus, the optimal value of Z is 1800.
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