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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 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$, x, y \geq 0$
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 $= Z = 2x + 3y$ 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 $O(0, 0), B_1(0, 5), G_1(2, 4), F_1(4, 2)$ and $E_1(4, 0).$
The values of $Z$ at these corner points are as follows.
Corner point
 $Z = 2x + 3y$
$O$ $0$
$_{B1}$ $15$
$G_1$ $16$
$F_1$ $14$
$E_1$ $8$
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 25 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 35 Marks
There are two types of fertilizers $F_{1 }$ and $F_2. F_{1 }$ consists of $10\%$ nitrogen and $6\%$ phosphoric acid and $​F_{2 }$ consists of $5\%$ nitrogen and $10\%$ phosphoric acid. After testing the soil conditions, a farmer finds the she needs atleast $14\ kg$ of nitrogen and $14\ kg$ of phosphoric acid for her crop. If $F_{1 }$ costs $Rs. 6/kg$ and $F_{2 }$ costs $Rs. 5/kg,$ determine how much of each type of fertilizer should be used so that the nutrient requirements are met at minimum cost. What is the minimum cost?
Answer
Suppose $x \ kg$ of fertilizer $F_1$ and and $y \ kg$ of fertilizer $F_2$ is used to meet the nutrient requirements.$F_1$ consists of $10\%$ nitrogen and $F_2$ consists of $5\%$ nitrogen.
But, the farmer needs atleast $14\ kg$ of nitorgen for the crops.
$10\%$ of $x \ kg + 5\%$ of $y \ kg 14\ kg$
$\Rightarrow\frac{\text{x}}{10}+\frac{\text{y}}{20}\geq14$
$\Rightarrow2\text{x}+\text{y}\geq280$
Similarly, $F_1$ consists of $6\%$ phosphoric acid and $F_2$ consists of $10\%$ phosphoric acid.
But, the farmer needs atleast $14\ kg$ of phosphoric acid for the crops.
$\Rightarrow\frac{6\text{x}}{100}+\frac{10\text{y}}{100}\geq14$
$\Rightarrow3\text{x}+5\text{y}\geq700$
The cost of fertilizer $F_1$ is $Rs. 6/kg$ and fertilizer $F_2$ is $Rs. 5/kg,$
Therefore, total cost of $x \ kg$ of fertilizer $F_1$ and and $y \ kg$ of fertilizer $F_2$ is $Rs. (6x + 5y).$
Thus, the given linear programming problem is
Minimise $Z = 6x + 5y$
Subject to the constraints
$2x + y \geq 280$
$3x + 5y \geq 700$
$x, y \geq 0$
The feasible region determined by the given constraints can be diagrammatically represented as,

The coordinates of the corner points of the feasible region are al $\text{A}\Big(\frac{700}{3},0\Big), B(100, 80)$ and $C(0, 280).$
The value of the objective function at these points are given in the following table.
$\text{Corner Point}$ $\text{Z} = 6\text{x} + 5\text{y}$
$\Big(\frac{700}{3},0\Big)$ $6\times\frac{700}{3}+5 \times 0= 1400$
$(100, 80)$ $6 \times 100+ 5 \times 80 = 1000 → \text{Minimum}$
$(0, 280)$ $6 \times 0 + 5 \times 280 =1400$
The smallest value of $Z$ is $1000$ which is obtained at $x = 100, y = 80.$
It can be seen that the open half$-$plane represented by has no common points with the feasible region.
So, the minimum value of $Z$ is $1000.$
Hence, $100\ kg$ of fertilizer $F_1$ and $80\ kg$ of fretilizer $F_2$ should be used so that the nutrient requirements are met at minimum cost.
The minimum cost is $Rs. 1,000.$
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Question 45 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 55 Marks
A manufacturer has three machine I, II, III installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must be operated for atleast 5 hours a day. She produces only two items M and N each requiring the use of all the three machines.
The number of hours required for producing 1 unit each of M and N on the three machines are given in the following table:
Item
Number of hours required on machines
 
I
II
III
M
1
2
1
N
2
1
1.25
She makes a profit of Rs. 600 and Rs. 400 on items M and N respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit?
Answer
Suppose x units of item M and y units of item N are produced to maximise the profit.Since each unit of item M require 1 hours on machine I and each unit of item N require 2 hours on machine I, therefore, the total hours required for producing x units of item M and yunits of item N on machine I are (2x + y).
But, machines I is capable of being operated for at most 12 hours.
2x + y ≤ 12 Similarly, each unit of item M require 2 hours on machine II and each unit of item N require 1 hour on machine II, therefore, the total hours required for producing x units of item M and yunits of item N on machine II are (x + 2y).
But, machines II is capable of being operated for at most 12 hours.
x + 2y ≤ 12 Also, each unit of item M require 1 hour on machine III and each unit of item N require 1.25 hour on machine III, therefore, the total hours required for producing x units of item M and yunits of item N on machine III are (x+ 1.25y).
But, machines III must be operated for atleast 5 hours.
x + 1.25 ≥ 5
The profit from each unit of item M is 2600 and each unit of item N is Rs. 400.
Therefore, the total profit from x units of item M and yunits of item N is Rs. (600x + 400y).
Thus, the given linear programming problem is Maximise Z = 600x + 400y
Subject to the constraints
2x + y ≤ 12
x + 2y ≤ 12
x + 1.25y ≥ 5
x, y ≥ 0
The feasible region determined by the given constraints can be diagrammatically represented.

The coordinates of the corner points of the feasible region are A(5, 0), B(6, 0), C(4, 4), D(0, 6) and E(0, 4).
The value of the objective function at these points are given in the following table.
Corner Point
Z = 600x + 400y
  
(5, 0)
600 × 5 + 400 × 0 = 3000
  
(6, 0)
600 × 6 + 400 × 0 = 3600
  
(4, 4)
600 × 4 + 400 × 4 = 4000
 → Maximum
(0, 6)
600 × 0 + 400 × 6 = 2400
  
(0, 4)
600 × 0 + 400 × 4 = 1600
  
The maximum value of Z is 4000 at x = 4, y = 4.
Hence, 4 units of item M and 4 units of item N should be produced to maximise the profit.
The maximum profit of the manufacturer is Rs. 4,000.
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Question 65 Marks
There are two types of fertilisers 'A' and 'B'. 'A' consists of 12% nitrogen and 5% phosphoric acid whereas 'B' consists of 4% nitrogen and 5% phosphoric acid. After testing the soil conditions, farmer finds that he needs at least 12kg of nitrogen and 12kg of phosphoric acid for his crops. If 'A' costs Rs. 10 per kg and 'B' cost Rs. 8 per kg, then graphically determine how much of each type of fertiliser should be used so that nutrient requiremnets are met at a minimum cost.
Answer
The given information can tabulated as follows:
Fertilizer
Nitrogen
Phosphoric Acid
Cost/kg (in Rs.)
A
12%
5%
10
B
4%
5%
8
 Let the requirement of fertilizer A by the farmer be x kg and that of B be y kg.
It is given that farmer requires atleast 12kg of nitrogen and 12kg of phosphoric acid for his crops.
The inequations thus formed based on the given information are as follows:
12100x + 4100y ≥ 12
⇒ 12x + 4y ≥ 1200
⇒ 3x + y ≥ 300 .....(1)
Also,
5x100 + 5y100 ≥ 12
⇒ 5x + 5y ≥ 1200
⇒ x + y ≥ 240 .....(2)
Total cost of the fertilizer Z = Rs. (10x + 8y)
Therefore, the mathematical formulation of the given linear programming problem can be stated as:
Minimize Z = 10x + 8y 
Subject to the constraints
3x + y ≥ 300 .....(1)
x + y ≥ 240 .....(2)
x ≥ 0, y ≥ 0 .....(3)
The feasible region determined by constraints (1) to (3) is graphically represented as:

Here, it is seen that the feasible region is unbounded.
The values of Z at the corner points of the feasible region are represented in tabular form as:
Corner point
Z = 10x + 8y
A(0, 300)
Z = 10 × 0 + 8 × 300 = 2400
B(30, 210)
Z = 10 × 30 + 8 × 210 =1980
C(240, 0)
Z = 7 × 240 + 8 × 0 = 2400
The open half plane determined by 10x + 8y < 1980 has no point in common with the feasible region.
So, the minimum value of Z is 1980.
The minimum value of Z is 1980, which is obtained at x = 30 and y = 210.
Thus, the minimum requirement of fertilizer of type A will be 30 kg and that of type B will be 210 kg.
Also, the total minimum cost of the fertilisers is Rs. 1980.
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Question 75 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 3$x + 2y = 210$ 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 85 Marks
A hospital dietician wishes to find the cheapest combination of two foods, $A$ and $B,$ that contains at least $0.5$ milligram of thiamin and at least $600$ calories. Each unit of $A$ contains $0.12$ milligram of thiamin and $100$ calories, while each unit of $B$ contains $0.10$ milligram of thiamin and $150$ calories. If each food costs $10$ paise per unit, how many units of each should be combined at a minimum cost?
Answer
Let required quantity of food $A$ and food $B$ bex and $y$ units.
Given, costs of one unit of food $A$ and $B$ are $10$ paise per unit each,
so costs ofx unit of food $A$ and $y$ unit of food $B$ are $10x$ and $10y$ respectively, let $z$ be total cost of foods,
so $Z = 10x + 10y$
Since one unit of food $A $and $B$ contain$ 0.12\ mg$ and $0.10\ mg$ of Thiamin respectively,
so, $x$ units of food $A$ and $y$ units of food $B$ contain $0.12\ xmg$ and $0.10\ ymg$ of Thiamin respectively but minimum requirement of Thiamin is $0.4\ mg,$
so$0.12\text{x}+10\text{y}\geq0.5$
$12\text{x}+10\text{y}\geq50$
$6\text{x}+5\text{y}\geq25 ($first constraint$)$
Since one unit of food $A$ and $B$ contain $100$ and $150$ calories respectively,
so $x$ units of food $A$ and $y$ units of food $B$ contain $100x$ and $150y$ units of Calories but minimum requirement of Calories is $600,$
so $100\text{x}+150\text{y}\geq600$
$\Rightarrow 2\text{x}+3\text{y}\geq12$ (second constraint)
Hence, mathematical formulation of $\text{LPP}$ is find $x$ and $y$ which
minimize $Z = 10x + 10y$
Subject to constraint,
$6\text{x}+5\text{y}\geq25$
$2\text{x}+3\text{y}\geq12$
$\text{x},\text{y}\geq0 [$Since quantity of food $A$ and $B$ can not be less than zero$]$
Region $6\text{x}+5\text{y}\geq25$:
$6x + 5y = 25$ meets axes at $\text{A}_1\Big(\frac{25}{6},0\Big), B_1(0, 5).$
Region not containing origin represents $6\text{x}+5\text{y}\geq25$ as $(0, 0)$ does not satisfy $6\text{x}+5\text{y}\geq25$.
Region $2\text{x}+3\text{y}\geq12$:
Line $2x + 3y = 12$ meets axes at $A_2(6, 0), B_2(0, 4).$
Region not containing origin represents $2\text{x}+3\text{y}\geq12$ as $(0, 0)$ does not satisfy $2\text{x}+3\text{y}\geq12$.
Region $\text{x},\text{y}\geq0$ represent first quadrant in $xy-$plane.

Unbounded shaded region $A_2 PB_1$ represents feasible region with corner points $A_2(6, 0),$ $\text{P}\Big(\frac{15}{8},\frac{11}{4}\Big), B_1(0, 5)$
The value of $Z = 10x + 10y$ at
$A_2(6, 0) = 10 (6) + 10(0) = 60$
$\text{P}\Big(\frac{15}{8},\frac{11}{4}\Big)=10\Big(\frac{15}{8}\Big)+10\Big(\frac{11}{4}\Big)=\frac{370}{8}=46\frac{1}{4}$
$B_1(0, 5) = 10(0) + 10(5) = 50$
Smallest value of $Z$ is $46\frac{1}{4}.$
Now open half plane $10\text{x}+10\text{y}<\frac{370}{8}$
$= 8x + 8y < 370$ has no point in common with feasible region,
so smallest value is the minimum value.
Hence, Required quantity of food $\text{A}=\frac{15}{8}$ units, food $\text{B}=\frac{11}{4}$ units
minimum cost $- Rs. 46.25$
$\Rightarrow\frac{15}{8}=1.875$ units of food $A$ and $\frac{11}{4}=2.75$ units of $B.$
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Question 95 Marks
If a young man drives his vehicle at 25 km/hr, he has to spend Rs. 2 per km on petrol. If he drives it at a faster speed of 40 km/hr, the petrol cost increases to Rs. 5 per km. He has Rs. 100 to spend on petrol and travel within one hour. Express this as an LPP and solve the same.
Answer
Let young man drives x km at a speed of 25 km/hr and y km at a speed of 40 km/hr.
Clearly, x, y ≥ 0
It is given that, he spends Rs. 2 per km if he drives at a speed of 25 km/hr and Rs 5 per km if he drives at a speed of 40 km/hr.
Therefore, money spent by him when he travelled x km and y km is Rs. 2x and Rs. 5y respectively.
It is given that he has a maximum of Rs. 100 to spend.
Thus, 2x + 5y ≤ 100
Time spent by him when travelling with a speed of 25 km/hr = x25hr
Time spent by him when travelling with a speed of 40 km/hr = x40hr
Also, the available time is of 1 hour.
x25 + y40 ≤ I = 40x + 25y ≤ 1000
The distance covered is Z = x + y which is to be maximised.
Max Z = x + y
Subject to
2x + 5y ≤ 100
40x + 25y ≤ 1000
x, y ≥ 0
First we will convert inequations into equations as follows:
2x + 5y = 100, 40x + 25y = 1000, x = 0 and y = 0
Region represented by 2x + 5y ≤ 100:
The line 2x + 5y = 100 meets the coordinate axes at A(50, 0) and B(0, 20) respectively.
By joining these points we obtain the line 2x + 5y = 100.
Clearly (0, 0) satisfies the 2x + 5y = 100.
So, the region which contains the origin represents the solution set of the inequation 2x + 5y ≤ 100.
Region represented by 40x + 25y ≤ 1000:
The line 40x + 25y = 1000 meets the coordinate axes at C(25, 0) and D(O, 40) respectively.
By joining these points we obtain the line 2x + y = 12.
Clearly (0, 0) satisfies the inequation 40x + 25y ≤ 1000.
So, the region which contains the origin represents the solution set of the inequation 40x + 25y ≤ 1000.
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 + 5y ≤ 100, 40x + 25y ≤ 1000, X ≥ 0, and y ≥ 0 are as follows.

The corner points are O(0, 0), B(0, 20), E(503, 403) and C(25, 0).
The values of Z at these corner points are as follows.
Corner point
Z = x + y
O
0
B
20
E
30
C
25
The maximum value of Z is 30 which is attained at E.
Thus, the maximum distance travelled by the young man is 30kms,
If he drives 503 km at a speed of 25 km/hr and 403km at a speed of 40 km/hr.
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Question 105 Marks
Tow godowns$, A$ and $B,$ have grain storage capacity of $100$ quintals and $50$ quintals respectively. They supply to $3$ ration shops, $D, E$ and $F,$ whose requirements are $60, 50$ and $40$ quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:
How should the supplies be transported in order that the transportation cost is minimum? 
Answer
Let godown $A$ supply $x$ quintals and $y$ quintals of grain to the shops $D$ and $E$ respectively.Then, $(100 - x - y)$ will be supplied to shop $F.$
The requirement at shop $D$ is $60$ quintals since, $x$ quintals are transported from godown $A.$
Therefore, the remaining $(60 - x)$ quintals will be transported from godown $B.$
Similarly, $(50 - y)$ quintals and $40 - (100 - x - y) i.e. (x + y - 60)$ quintals will be transported from godown $B$ to shop $E$ and $F$ respectively.
The given problem can be represented diagrammatically as follows.

Quantity of the grain cannot be negative.
Therefore, $x \geq 0, y \geq 0,$ and $100 - x - y \geq 0$
$x \geq 0, y \geq 0,$ and $x + y \leq 100$
$60 - x \geq 0, 50 - y \geq 0,$ and $x + y - 60 \geq 0$
$x \leq 60, y \leq 50$, and $x + y \geq 60$
Total transportation cost $Z$ is given by,
$Z = 6x + 3y + 2.5 (100 - x - y) + 4(60 - x) + 2(50 - y) + 3(x + y - 60)$
$= 6x + 3y + 250 - 2.5x - 2. 5y + 240 - 4x + 100 - 2y + 3x + 3y - 180$
$= 2.5x + 1. 5y + 410$
The given problem can be formulated as:
Minimize $Z = 2.5x + 1.5y + 410$
Subject to the constraints,
$x + y \leq 100$
$X \leq 60$
$y \leq 50$
$x + y \geq 60$
$x, y \geq 0$
First we will convert inequations into equations as follows:
$x + y = 100, x = 60, y = 50, x + y = 60, x = 0$ and $y = 0$
Region represented by $x + y \leq 100:$
The line $x + y = 100$ meets the coordinate axes at $A_1(100, 0)$ and $B_1(0, 100)$ respectively.
By joining these points we obtain the line $x + y = 100.$
Clearly $(0, 0)$ satisfies the $x + y \leq 100.$
So, the region which contains the origin represents the solution set of the inequation
$x + y \leq 100.$
Region represented by $x \leq 60:$
$x = 60$ is the line that passes $(60, 0)$ and is parallel to the $y-$axis.
The region to the left of the line $x = 60$ will satisfy the inequation $x \leq 60.$
Region represented by $y \leq 50:$
$y = 50$ is the line that passes $(0, 50)$ and is parallel to the $x-$axis.
The region below the line $y = 50$ will satisfy the inequation $y \leq 50.$
Region represented by $x + y \geq 60:$
The line $x + y= 60$ meets the coordinate axes at $C_1(60, 0)$ and $D_1(0, 60)$ respectively.
By joining these points we obtain the line $x + y = 60.$
Clearly $(0, 0)$ does not satisfies the inequation $x + y \geq 60.$
So, the region which does not contain the origin represents the solution set of the inequation $x + y \geq 60.$
Region represented by $x \geq 0$ and $y \geq 0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $x \geq 0,$ and $y \geq 0.$
The feasible region determined by the system of constraints $x + y \leq 100, x \leq 60, y \leq 50, x + y \geq 60, x \geq 0$ and $y \geq 0$ are as follows.

The corner points are $C_1(60, 0), G_1(60, 40), F_1(50, 50),$ and $E_1(10, 50).$
The values of $Z$ at these corner points are as follows:​​​​​​​
Corner point
 $Z = 2.5x + 1.5y + 410$
$C_1(60, 0)$ $560$
$G_1(60, 0)$ $620$
$F_1(50, 5)$ $610$
$E_1(10, 0)$ $510$
The minimum value of $Z$ is $510$ at $E_1(10, 50).$
Thus, the amount of grain transported from $A$ to $D, E,$ and $F$ is $10$ quintals, $50$ quintals, and $40$ quintals respectively and from $B$ to $D, E,$ and $F$ is $50$ quintals, $0$ quintals, and $0$ quintals respectively.
The minimum cost is $Rs. 510.$
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Question 115 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 125 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 $\text{LPP}$ does not exist.
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Question 135 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 145 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 155 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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Question 165 Marks
Maximize $Z = x + y$
Subject to $-2\text{x}+\text{y}\leq1$
$\text{x}\leq2$
$\text{x}+\text{y}\leq3$
$\text{x},\text{y}\geq0$
Answer
Converting the given inequations into equation,
$-2x + y = 1, x = 2, x + y = 3, x= y = 0$

Region represented by $- 2x + y = 1$:
Line $- 2x + y = 1$ meets coordinate axes at $\text{A}_1\Big(\frac{-1}{2},0\Big)$ and $B_1(0, 1),$ clearly $, (0, 0)$ satisfies $-2\text{x}+\text{y}\leq1,$ so region containing origin represents $-2\text{x}+\text{y}\leq1$ in $xy-$ plane.
Region represented by $\text{x}\leq2$:
Linex $- 2$ is parallel to $y-$ axis and meets $x-$ axis at $A_3(2, 0)$.
Clearly $,(0, 0)$ satisfies $\text{x}\leq2,$ so region containing origin represents $\text{x}\leq2$ in $xy-$ plane.
Region represented by $\text{x}+\text{y}\leq3$:
Line $x + y - 3$ meets coordinate axes at $A_2(3, 0)$ and $B_2(0, 3)$.
Clearly $, (0, 0)$ satisfies $\text{x}+\text{y}\leq3,$ so region containing origin represents $\text{x}+\text{y}\leq3$ in $xy-$ plane.
Region represented $\text{x},\text{y}\geq0$:
It represents first quadrant in $xy-$ plane.
So, shaded region $OA_3 \ PQ8,$ represents the feasible region.
Coordinates of $P(2, 1)$ is obtained by solving $x + y = 3$ and $x = 2, \text{Q}\Big(\frac{2}{3},\frac{7}{3}\Big)$ by solving $-2x + y = 1$ and $x + y = 3$.
The value of $Z = x + y $ at
$\text{O}(0, 0) = 0 + 0 = 0$
$\text{A}_3(2, 0) = 2 + 0 = 2$
$\text{P}(2, 1) = 2 +1 = 2$
$\text{Q}\Big(\frac{2}{3},\frac{7}{3}\Big)=\frac{2}{3}+\frac{7}{3}=3$
$\text{B}_1(0, 1) = 0 + 1 = 1$
So, maximum $Z = 3$ is at every point on the line joining $PQ$.
Hence, maximum $z = 3 $ at $x = 2$ and $y = 1$ Or $\text{x}=\frac{2}{3}$ and $\text{y}=\frac{7}{3}.$
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Question 175 Marks
Find the minimum value of 3x + 5y subject to the constraints:
$-2\text{x}+\text{y}\leq4,\text{x}+\text{y}\geq3,$ $\text{x}-2\text{y}\leq2,\text{x},\text{y}\geq0.$
Answer
First, we will convert the given inequations into equations, we obtain the following equations:
-2x + y = 4, x + y = 3, x - 2y = 2, x = 0 and y = 0.
The line -2x + y = 4 meets the coordinate axis at A(-2, 0) and B(0, 4).
Join these points to obtain the line -2x + y = 4.
Clearly, (0, 0) satisfies the inequation $-2\text{x}+\text{y}\leq4$.
So, the region in xy-plane that contains the origin represents the solution set of the given equation.
The line x + y = 3 meets the coordinate axis at C(3, 0) and D(0, 3).
Join these points to obtain the line x + y = 3.
Clearly, (0, 0) does not satisfies the inequation $\text{x}+\text{y}\geq3$.
So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
The line x - 2y = 2 meets the coordinate axis at E(2, 0) and F(0, -1).
Join these points to obtain the line x - 2y = 2.
Clearly, (0, 0) satisfies the inequation $\text{x}-2\text{y}\leq2$.
So, the region in xy-plane that 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 cornerpoint of the feasible region are B(0, 4), D(0, 3) and $\text{G}\Big(\frac{8}{3},\frac{1}{3}\Big)$.
The values of Z at these corner points are as follows.
$\text{Corner point}$
$\text{Z}=3\text{x}+5\text{y}$
$\text{B}(0, 4)$
$3\times0+5\times4=20$
$\text{D}(0, 3)$
$3\times0+5\times3=15$
$\text{G}\Big(\frac{8}{3},\frac{1}{3}\Big)$
$3\times\frac{8}{3}+5\times\frac{1}{3}=\frac{29}{3}$
We see that minimum value of the objective function Z is $\frac{29}{3}$ which is at $\text{G}\Big(\frac{8}{3},\frac{1}{3}\Big)$.
Thus, the optimal value of Z is $\frac{29}{3}$.
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Question 185 Marks
Find graphically, the maximum value of Z = 2x + 5y, subject to constraints given below:
$2\text{x}+4\text{y}\leq8$
$3\text{x}+\text{y}\leq6$
$\text{x}+\text{y}\leq4$
$\text{x}\geq0,\text{y}\geq0$
Answer
Converting the inequations into equations, ew obtain the lines.
2x + 4y = 8, 3x + y = 6, x + y =4, x = 0, y = 0.
These lines are drawn on a suitable scale and the feasible region of the LPP is shaded in the graph.

From the graph we can see the corner point as (0, 2) and (2, 0).
Now, solving the equations 3x + y = 6 and 2x +4y = 8 we get the values of x and y as $\text{x}=\frac{8}{5}$ and $\text{y}=\frac{6}{5}$.
Substituting $\text{x}=\frac{8}{5}$ and $\text{y}=\frac{6}{5}$ in Z = 2x + 5y we get,
$\text{z}=2\Big(\frac{8}{5}\Big)+5\Big(\frac{6}{5}\Big)$
$\text{z}=\frac{46}{5}$
Hence maximum value of Z is $\frac{46}{5}$ at $\text{x}=\frac{8}{5}$ and $\text{y}=\frac{6}{5}$.
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Question 195 Marks
Solve the following LPP graphically:
Manimize Z = 6x + 3y
Subject to the constraints:
$4\text{x}+\text{y}\geq80$
$\text{x}+5\text{y}\geq115$
$3\text{x}+2\text{y}\leq150$
$\text{x}\geq0,\text{y}\geq0$
Answer
The given contraints are:
$4\text{x}+\text{y}\geq80$
$\text{x}+5\text{y}\geq115$
$3\text{x}+2\text{y}\leq150$
$\text{x}\geq0,\text{y}\geq0$
Converting the given inequation into equation, we get
4x + y = 80, x + 5y = 115, 3x + 2y = 150, x = 0 and y = 0
These lines are drawn on the graph and the shaded region ABC represents the feasible region of the given LPP.

It can be observed that the feasible region is bounded.
The coordinates of the corner points of the feasible region are A(2, 72), B(15, 20), and C(40, 15).
The values of the objective function, Z at these corner points are given in the following table:
Corner point Value of the objective Function Z = 6x + 3y
A(2, 72) Z = 6 × 2 + 3 × 72 = 228
B(15, 20) Z = 6 × 15  + 3 × 20 = 150
C(40, 15) Z = 6 × 40 + 3 × 15 = 150
From the table, Z is manimum at x = 15 and y = 20 and the manimum value of Z is 150.
Thus, the manimum value of Z is 150.
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Question 205 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 215 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 225 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 235 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 245 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 255 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, $x \geq 0, y \geq 0$
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 $\text{LPP}$ 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 $24$
$x \geq 0, y \geq 0$
First we will convert inequations into equations as follows:
$1.5x + 3y = 42, 3x + y = 24, x = 0$ and $y = 0$
Region represented by $1.5x + 3y \leq 42:$
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 $1.5x + 3y \leq 42.$
Region represented by $3x + y \leq 24:$
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 $3x + y \leq 24.$
So the region which contains the origin represents the solution set of the inequation $3x + y \leq 24$.
Region represented by $x \geq 0$ and $y \geq 0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $x \geq 0$ and $y \geq 0.$
The feasible region determined by the system of constraints $1.5x + 3y \leq 42, 3x + y \leq 24, x \geq 0$ and $y \geq 0$ 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 265 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 275 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 285 Marks
If a young man drives his scooter at a speed of 25km/hr, he has to spend Rs. 2 per km on petrol. If he drives the scooter at a speed of 40km/hr, it  produces air pollution and increases his expenditure on petrol to Rs. 5 per km. He has a maximum of Rs. 100 to spend on petrol and travel a maximum distance in one hour time with less polution. Express this problem as an LPP and solve it graphically. What value do you find here.
Answer
Let the distance covered with speed 25 km/hr = x km
And, the distance covered with speed 40 km/hr = y km
We know that,
$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$
Thus, Maximum speed covered within one hour $= \frac{\text{x}}{25} + \frac{\text{y}}{40} \leq 1$
Thus, According to question,
Maximum speed covered, $\text{Z}_\text{max} = \text{x} + \text{y}$
Subject to the constrains, $\text{2x + 5y} \leq 100$
$\frac{\text{x}}{25} + \frac{\text{y}}{40} \leq 1$
$\text{x} , \text{y} \geq 0$
Now plotting both line on graph paper, we have,

from graph, OABC is feasible region.
Thus, maximum distance covered  $= \frac{50}{3} + \frac{40}{3} = 30 \text{km}$ 
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Question 295 Marks
A small firm manufactures gold rings and chains. The total number of rings and chains manufactured per day is at most $24$. It takes $1$ hour to make a ring and $30$ minutes to make a chain. The maximum number of hours available per day is $16$. If the profit on a ring is $Rs. 300$ and that on a chain is $Rs. 190,$ find the number of rings and chains that should be manufactured per day, so as to earn the maximum profit. Make it as an $\text{LPP}$ and solve it graphically.
Answer
Let required number of gold rings and chains are $x$ and $y$ respectively.
Since,profits on each ring and chains are $Rs. 300$ and $Rs. 190$ respectively, so, profit on $x$ units of ring and $y$ units of chains are $Rs. 300x$ and $Rs. 190y$ respectively
Let $Z$ be total profit so
$Z = 300x + 190y$
Since each unit of ring and chain require $1$ hr and $30$ min. to make respectively, so,
$X$ units of rings and y units of rings require $60x$ and $30y$ min. to make respectivley, but total time available to make is $16 \times 60 = 960,$ so,
$60x + 30y \leq 960$
$= 2x + y \leq 32  \ ($first constraint$)$
Given, total number of rings and chains manufactured is at most $24,$ so,
$x + y \leq 24 \  ($second constraint$)$
Hence, mathematical formulation of $\text{LPP}$ is find $x$ and $y$ which
Maximize $Z = 300x + 160y$
Subject to constriants,
$2x + y \leq 32$
$x + y \leq 4$
$x, y \geq 0  \ [$Since production can not be less than zero$]$
Region $2x + y \leq 32:$
Line $2x + y = 32$ meets axes at $A_1(16, 0), B_1(0, 32)$ respectively.
Region containing origin represents $2x + y \leq 32$ as $(0, 0)$ satisfies $2x +y \leq 32$
Region $x + y \leq 24:$
Line $x + y = 24$ meets axes at $A_2(24, 0), B_2(0, 24)$ respectively.
Region containing origin represents $x + y \leq 24 as (0, 0)$ satisfies $x + y \leq 24$.
Region $x, y \geq 0:$
It represent first quandrant
Shaded region $OA_1\ PB_2$ represents feasible region.
Point $P(8, 16)$ is obtained by solving $2x + y = 32$ and $x + y = 24$.

The value of $Z = 300x + 160y$ at
$O(0, 0) = 300(0) + 160(0) = 0$
$A_1(16, 0) = 300(16) + 160(0) = 4800$
$P(8, 16) = 300(8) + 160(16) = 4960$
$B_2(0, 24) = 300(0) + 160(24) = 3840$
Maximum $Z = 4960$ at $ x = 8, y = 16$
Number of rings $= 8,$ chains $= 16$
Maximum profit $= Rs. 4960$
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Question 305 Marks
Maximize Z = 50x + 30y
Subject to
$2\text{x}+\text{y}\leq18$
$3\text{x}+2\text{y}\leq34$
$\text{x},\text{y}\geq0$
Answer
First, we will convert the given inequations into equations, we obtain the following equations:
2x + y = 18, 3x + 2y = 34
Region represented by 2x + y ≥ 18:
The line 2x + y = 18 meets the coordinate axes at A(9, 0) and B(0, 18) respectively. By joining these points we obtain the line 2x + y = 18.
Clearly (0,0) does not satisfies the inequation 2x + y ≥ 18.
So,the region in xy plane which does not contain the origin represents the solution set of the inequation 2x + y ≥ 18.
Region represented by 3x + 2y ≤ 34:
The line 3x + 2y = 34 meets the coordinate axes at $\text{C}\Big(\frac{34}{3},0\Big)$ and D(0, 17) respectively. By joining these points we abtain the line 3x + 2y = 34.
Clearly (0. 0) satisfies the inequation 3x + 2y ≤ 34. So, the region containing the origin represents the silution set of the inequation 3x + 2y ≤ 34.
The corner of the feasible region are A(9, 0), $\text{C}\Big(\frac{34}{3},0\Big)$ and E(2, 14).

The values of Z at these corner points are as follows.
$\text{Corner point}$
$\text{Z}=50\text{x}+30\text{y}$
$\text{A}(9, 0)$
$50\times9+3\times0=450$
$\text{C}\Big(\frac{34}{3},0\Big)$
$50\times\frac{34}{3}+30\times0=\frac{1700}{3}$
$\text{E}(2, 14)$
$50\times2+30\times14=520$
Therefore, the maximum value of Z $\frac{1700}{3}$ is at the point $\Big(\frac{34}{3},0\Big)$.
Hence, $\text{x}=\frac{34}{3}$ and $\text{y}=0$ is the optimal solution of the given LPP.
Thus, the optimal value of Z is $\frac{1700}{3}$.
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Question 315 Marks
An automobile manufacturer makes automobiles and trucks in a factory that is divided into two shops. Shop A, which performs the basic assembly operation, must work 5 man-days on each truck but only 2 man-days on each automobile. Shop B, which performs finishing operations, must work 3 man-days for each automobile or truck that it produces. Because of men and machine limitations, shop A has 180 man-days per week available while shop B has 135 man-days per week. If the manufacturer makes a profit of Rs 30000 on each truck and Rs 2000 on each automobile, how many of each should he produce to maximize his profit? Formulate this as a LPP.
Answer
Let x number of trucks and y number of automobiles were produced to maximize the profit.
Since, the manufacturer makes profit of Rs. 30000 on each truck and Rs. 2000 on each automobile.
Therefore, on x number of trucks and y number of automobiles profit would be Rs. 30000x and Rs. 2000yrespectively.
Total profit = Rs. (30000x + 2000y)
​Let Z denote the total profit
Then, Z = 30000x + 2000y
Since, 5 man-days and 2 man-days were required to produce each truck and automobile at shop A.
Therefore, 5x man-days and 2y man-days are required to produce x trucks  and y automobiles at shop A.
Also,
Since 3 man-days were required to produce each truck and automobile at shop B.
Therefore, 3x man-days and 3y man-days are required to produce x trucks and y automobiles​.
As, shop A has 180 man-days per week available while shop B has 135 man-days per week.
$\therefore5\text{x}+2\text{y}\leq180,3\text{x}+3\text{y}\leq135$
Number of trucks and automobiles cannot be negative.
$\therefore\text{x},\text{y}\geq0$
Hence, the required LPP is as follows:
Maximize Z = 30000x + 2000y
Subject to
$5\text{x}+2\text{y}\leq180,$
$3\text{x}+3\text{y}\leq135,$
$\text{x}\geq0,\text{y}\geq0$
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Question 325 Marks
A cooperative society of farmers has $50$ hectares of land to grow two crops $X$ and $Y.$ The profits from crops $X$ and $Y$ per hectare are estimated as $Rs. 10,500$ and $Rs. 9,000$ respectively. To control weeds, a liquid herbicide has to be used for crops $X$ and $Y$ at the rate of $20$ litres and $10$ litres per hectare, respectively. Further not more than $800$ litres of herbicide should be used in order to protect fish and wildlife using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total profit of the society $? $
Answer
Let $x$ hectores of land grows crop $X.$
Let $y$ hectores of land grows crop $Y.$
Then the mathematical model of the $LPP$ is as follows:
Maximize $Z = 10,500x + 9,000$
Subject to $x + y <50, 20x + 10y$
$5800$ and $x 20, y20$
To solve the $LPP$ we draw the lines, $x + y = 50, 20x + 10y = 800$
The feasible region of the $LPP$ is shaded in graph
 
The coordinates of the vertices $($Corner $-$ points$)$ of shaded feasible region $ABC$ are $A(40, 0), B(30, 20)$ and $C(0, 50).$
The values of the objective of function at these points are given in the following table:
Point $(x_1, x_2)$
Value of objective function $= Z = 10,500x + 9,000y$
$A(40, 0)$ $Z = 4,20,000$
$B(30, 20)$ $Z = 4,95,000$
$C(0, 50)$ $Z = 4,50,000$
$30$ hectors of land should be allocated to crop $X$ and
$20$ hectors of land should be allocated to crop $Y$ to maximize the profit
The maximum profit that can be eared is $Rs. 4,95,000.$
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Question 335 Marks
An oil company has two depots$, A$ and $B,$ with capacities of $7000$ litres and $4000$ litres respectively. The company is to supply oil to three petrol pumps, $\text{D, E, F}$ whose requirements are $4500, 3000$ and $3500$ litres respectively. The distance $($in $\ km)$ between the depots and petrol pumps is given in the following table:

Assuming that the transportation cost per km is $Rs. 1.00$ per litre, how should the delivery be scheduled in order that the transportation cost is minimum?
Answer
Let $x$ and y litres of oil be supplied from $A$ to the petrol pumps $D$ and $E.$
Then, $(7000 − x − y) L$ will be supplied from $A$ to petrol pump $F.$
The requirement at petrol pump $D$ is $4500 L.$
Since$, x L$ are transported from depot $A,$ the remaining $(4500 − x) L$ will be transported from petrol pump $B.$
Similarly, $(3000 − y) L$ and $[3500 − (7000 − x − y)]$
L i.e. $(x + y − 3500) L$ will be transported from depot $B$ to petrol pump $E$ and $F.$ respectively.
The given problem can be represented diagrammatically as follows.

Since, quantity of oil are non-negative quantities.
Therefore,
$x \geq 0, y \geq 0$ and $(7000 - x - y) \geq 0$
$\Rightarrow x \geq 0, y \geq 0$ and $x + y \leq 7000$
$4500 - x \geq 0, 3000 - y \geq 0$ and $x + y - 3500 \geq 0$
$\Rightarrow x \leq 4500, y \leq 3000$ and $x + y \geq 3500$
Cost of transporting $10 L$ of petrol $= RS. 1$
Cost of transporting $1 L$ of petrol $= Rs. 110$
Therefore, total transportation cost is given by,
$\text{Z}=\frac{7}{10}\times\text{x}+\frac{6}{10}\text{y}+\frac{3}{10}(7000-\text{x}-\text{y})+\frac{3}{10}(4500-\text{x})$
$+\frac{4}{10}(3000-\text{y})+\frac{2}{10}(\text{x}+\text{y}-3500)$
$= 0.3x + 0.1y + 3950$
The problem can be formulated as follows.
Minimize $Z = 0.3x + 0.1y + 3950$
Subject to the constraints,
$x + y \leq 7000$
$x \leq 4500$
$y \leq 3000$
$x + y \geq 3500$
$x, y \geq 0$
First we will convert inequations into equations as follows:
$x + y = 7000, x = 4500, y = 3000, x + y = 3500, x = 0$ and $y = 0$
Region represented by $x + y \leq 7000:$
The line $x + y = 7000$ meets the coordinate axes at $A_1(7000, 0)$ and $B_1(0, 7000)$ respectively.
By joining these points we obtain the line $x + y = 7000.$
Clearly $(0, 0)$ satisfies the $x + y = 7000.$
So, the region which contains the origin represents the solution set of the inequation $x + y \leq 7000.$
Region represented by $x ​ \leq 4500:$
The line ​$x = 4500$ is the line passes through $C_1(4500, 0)$ and is parallel to $Y$ axis.
The region to the left of the line $x = 4500$ will satisfy the inequation $x ​ \leq 4500.$
Region represented by $y ​ \leq 3000:$
The line $​y = 3000$ is the line passes through $D_1(0, 3000)$ and is parallel to $X$ axis.
The region below the line $y = 3000$ will satisfy the inequation $y ​ \leq 3000.$
Region represented by $x + y \geq 3500:$
The line $x + y = 7000$ meets the coordinate axes at $E_1(3500, 0)$ and $F_1(0, 3500)$ respectively.
By joining these points we obtain the line $x + y = 3500.$
Clearly $(0, 0)$ satisfies the $x + y = 3500.$
So, the region which contains the origin represents the solution set of the inequation $x + y \geq 3500.$
Region represented by $x \geq 0$ and $y \geq 0:$
Since, every point in the first quadrant satisfies these inequations.
So, the first quadrant is the region represented by the inequations $x \geq 0,$ and $y \geq 0.$
The feasible region determined by the system of constraints $x + y \leq 7000, x ​ \leq 4500, y ​ \leq 3000, x + y \geq 3500, x \geq 0$ and $y \geq 0$ are as follows.
$\text{GRAPH}$
The corner points of the feasible region are $E_1(3500, 0), C_1(4500, 0), I_1(4500, 2500), H_1(4000, 3000),$ and $G_1(500, 3000).$
The values of $Z$ at these corner points are as follows.
Corner point
$Z = 0.3 + 0.1y + 3950$
$E_1(3500, 0)$ $5000$
$C_1(4500, 0)$ $5300$
$I_1(4500, 2500)$ $5550$
$H_1(4000, 3000)$ $5450$
$G_1(500, 3000)$ $4400$
The minimum value of $Z$ is $4400$ at $G_1(500, 3000).$
Thus, the oil supplied from depot $A$ is $500 L, 3000 L,$ and $3500 L$ and from depot $B$ is $4000 L, 0 L,$ and $0 L$ to petrol pumps $D, E,$ and $F$ respectively.
The minimum transportation cost is $Rs. 4400.$
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Question 345 Marks
A rubber company is engaged in producing three types of tyres A, B and C. Each type requires processing in two plants, Plant I and Plant II. The capacities of the two plants, in number of tyres per day, are as follows:
Plant
A
B
C
I
50
100
100
II
60
60
200
The monthly demand for tyre A, B and C is 2500, 3000 and 7000 respectively. If plant I costs Rs. 2500 per day, and plant II costs Rs. 3500 per day to operate, how many days should each be run per month to minimize cost while meeting the demand? Formulate the problem as LPP.
Answer
Let plant I be run for x days and plant II be run for y dayes
Then,
Tyres
Plant I (4)
Plant II (y)
Demand
 
A
50
60
2500
B
100
60
3000
C
100
200
7000
Minimum demand for Tyres A,B and C is 2500, 3000 and 7000 respectively. The demand can be more then the minimum demand.
Therefore, the incquations will be.
$50\text{x}+60\text{y}\geq2500$
$100\text{x}+60\text{y}\geq3000$
$100\text{x}+200\text{y}\geq7000$
Also, the objective function is Z = 2500x + 3500y
Hence, the required LPP is as follows:
Minimise Z = 2500x + 3500y
subject to
$50\text{x}+60\text{y}\geq2500$
$100\text{x}+60\text{y}\geq3000$
$100\text{x}+200\text{y}\geq7000$
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Question 355 Marks
Maximum Z = 2x + 3y
Subject to
$\text{x}+\text{y}\geq1$
$10\text{x}+\text{y}\geq5$
$\text{x}+10\text{y}\geq1$
$\text{x},\text{y}\geq0$
Answer
First, we will convert the given inequations into equations, we obtain the following equations:
x + y = 1, 10x +y = 5, x + 10y = 1, 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) does not satisfies the inequation x + y ≥ 1.
So, the region in xy plane which does not contain the origin represents the solution set of the inequation x + y ≥ 1.
Region represented by 10x + y ≥ 5:
The line 10x + y = 5 meets the coordinate axes at $\text{C}\Big(\frac{1}{2},0\Big)$ and D(0, 5) respectively.
By joining these points we obtain the line 10x + y = 5.
Clearly (0, 0) does not satisfies the inequation 10x + y ≥ 5.
So, the region which does not contains the origin represents the solution set of the inequation 10x +y ≥ 5.
Region represented by x + 10y ≥ 1:
The line x + 10y = 1 meets the coordinate axes at A(1, 0) and $\text{F}\Big(0,\frac{1}{2}\Big)$ respectively.
By joining these points we obtain the line.
x + 10y = 1.
Clearly (0, 0) does not satisfies the inequation x + 10y ≥ 1.
So, the region which does not contains the origin represents the solution set of the inequation x + 10y ≥ 1.
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, 10x + y ≥ 5, x + 10y ≥ 1, x ≥ 0, and y ≥ 0, are as follows.

The feasible region is unbounded.
Therefore, the maximum value is infinity i.e. the solution is unbounded.
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Question 365 Marks
A manufacturer can produce two products, A and B, during a given time period. Each of these products requires four different manufacturing operations: grinding, turning, assembling and testing. The manufacturing requirements in hours per unit of products A and B are given below.
 
A
B
Grinding
1
2
Turning
3
1
Assembling
6
3
Testing
5
4
The available capacities of these operations in hours for the given time period are: grinding 30; turning 60, assembling 200; testing 200. The contribution to profit is Rs 20 for each unit of A and Rs 30 for each unit of B. The firm can sell all that it produces at the prevailing market price. Determine the optimum amount of A and B to produce during the given time period. Formulate this as a LPP.
Answer
Given inform ation can be tabulated:-
Product
Grinding
Turning
Assembling
Testing
Profit
A
1
3
6
5
2
B
2
1
3
4
3
Maximum
capacity
30 hours
60 hours
200 hours
200 hours
  
Let required production of product A and B be x and y respectively
Given, profits on one unit of product A and B are Rs. 2 and Rs. 3 respectively, so profits on x units of product A and y units of product B are given by 2x and 3y respectively. Let Z be total profit, so
Z = 2x + 3y
Given, production of 1 unit of product A and B require 1 hour and 2 hours of grinding respectively, so, production of x units of product A and y units of of product B require x hours and 2y hours of grinding respectively but maximum time available for grinding is 3 hours, so
$\text{x}+2\text{y}\leq30$ (First constraint)
Given, production of 1 unit of product A and B require 3 hours and 1 hours of turning respectively, so x units of product A and y units of of product B require 3x hours and y hours of turning respectively but total time available for turning is 60 hours, so
$3\text{x}+\text{y}\leq60$ (Second constraint)
Given, production of 1 unit of product A and B require 6 hour and 3 hours of assembling respectively, so productinon of x units of product A and y units of product 8 require 6x hours and 3y hours of assembling respectively but total time available for assembling is 200 hours, so
$6\text{x}+3\text{y}\leq200$ (Third constraint)
Given, production of 1 unit of product A and B require 5 hours and 4 hours of testing respectively, so productinon of x units of product A and y units of product B require 5x hours and 4y hours of testing respectively but total time available for testing is 200 hours, so
$5\text{x}+4\text{y}\leq200$ (Fourth constraint)
Hence, mathematical formulation of LPP is,
Find x and y which
maximize Z = 2x + 3y
Subject to constraints,
$\text{x}+2\text{y}\leq30$
$3\text{x}+\text{y}\leq60$
$6\text{x}+3\text{y}\leq200$
$5\text{x}+4\text{y}\leq200$
and, $\text{x},\text{y}\geq0$ [Since production can not be negative].
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Question 375 Marks
A manufacturer has three machines installed in his factory. machines $I$ and $II$ are capable of being operated for at most $12$ hours whereas Machine $III$ must operate at least for $5$ hours a day. He produces only two items, each requiring the use of three machines. The number of hours required for producing one unit each of the items on the three machines is given in the following table:
Item Number of hours required by the machine
  $I$ $II$ $III$
$A$ $1$ $2$ $1$
$B$ $2$ $1$ $\frac{5}{4}$
He makes a profit of $Rs. 6.00$ on item $A$ and $Rs. 4.00$ on item $B.$ Assuming that he can sell all that he produces, how many of each item should he produces so as to maximize his profit? Determine his maximum profit. Formulate this $\text{LPP}$ mathematically and then solve it.
Answer
Let $x$ units of item $A$ and $Y$ units of item $B$ be manufactured. Therefore, As we are given,
Item Number of hours required by the machine
  $I$ $II$ $III$
$A$ $1$ $2$ $1$
$B$ $2$ $1$ $\frac{5}{4}$
Machines $I$ and $II$ are capable of being operated for at most $12$ hours whereas Machine $Ill$ must operate at least for $5$ hours a day.
According to question, the constraints are
$\text{x}+2\text{y}\leq12$
$2\text{x}+\text{y}\leq12$
$\text{x}+\frac{5}{4}\text{y}\geq5$
He makes a profit of $Rs. 6.00$ on item $A$ and $Rs. 4.00$ on item $B.$
Profit made by him in producing $x$ items of $A$ and $y$ items of $B$ is $6x + 4y.$
Total profit $Z = 6x + 4y$ which is to be maximised.
Thus, the mathematical formulation of the given linear programmimg problem is
Max $Z = 6x + 4y$
Subject to
$\text{x}+2\text{y}\leq12$
$2\text{x}+\text{y}\leq12$
$\text{x}+\frac{5}{4}\text{y}\geq5$
$\text{x},\text{y}\geq0$
First we will convert inequations into equations as follows:
$x + 2y = 12$
$2x + y= 12,$
$\text{x}+\frac{5}{4}\text{y}=5$
$x = 0$ and $y = 0$
Region represented by $\text{x}+2\text{y}\leq12$:
The line $x + 2y = 12$ meets the coordinate axes at $A_1(12, 0)$ and $B_1(0, 6)$ respectively.
By joining these points we obtain the line $x + 2y = 12.$
Clearly $(0, 0)$ satisfies the $x + 2y = 12.$
So, the region which contains the origin represents the solution set of the inequation $\text{x}+2\text{y}\leq12$.
Region represented by $2\text{x}+\text{y}\leq12$:
The line $2x + y = 12$ meets the coordinate axes at $C_1(6, 0)$ and $D_1(0, 12)$ respectively.
By joining these points we obtain the line $2x + y = 12.$
Clearly $(0, 0)$ satisfies the inequation $2\text{x}+\text{y}\leq12$.
So, the region which contains the origin represents the solution set of the inequation $2\text{x}+\text{y}\leq12$.
Region represented by $\text{x}+\frac{5}{4}\text{y}\geq5$:
The line $\text{x}+\frac{5}{4}\text{y}\geq5$ meets the coordinate axes at $E_1(5, 0)$ and $F_1(0, 4)$ respectively.
By joining these points we obtain the line $x + 2y = 5.$
Clearly $(0, 0)$ does not satisfies the inequation $\text{x}+\frac{5}{4}\text{y}\geq5$.
So,the region which does not contains the origin represents the solution set of the inequation.
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 $\text{x}+2\text{y}\leq12,2\text{x}+\text{y}\leq12,\text{x}+\frac{5}{4}\text{y}\geq5,\text{x}\geq0$ and $\text{y}\geq0$ are as follows.

Thus, the maximum profit is $Rs. 40$ obtained when $4$ units each of item $A$ and $B$ are manufactured.
The corner points are $B(0, 6), G(4, 4), C(6, 0), E(5, 0)$ and$ F(0, 4).$
The values of $Z$ at these corner points are as follows:
Corner point
 $Z = 6x + 4y$
$B_1$ $24$
$G_1$ $40$
$C_1$ $36$
$E_1$ $30$
$F_1$ $16$
The maximum value of $Z$ is $40$ which is attained at $G (4, 4).$
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Question 385 Marks
Maximum Z = 3x + 4y
Subject to
$2\text{x}+2\text{y}\leq80$
$2\text{x}+4\text{y}\leq120$
Answer
We have to maximize Z = 3x + 4y First, we will convert the given inequations into equations, we obtain the following equations: 2x + 2y = 80, 2x + 4y = 120 Region represented by 2x + 2y ≤ 80: The line 2x + 2y = 80 meets the coordinate axes at A(40, 0) and B(0, 40) respectively. By joining these points we obtain the line 2x + 2y = 80. Clearly (0, 0) satisfies the inequation 2x + 2y ≤ 80. So,the region containing the origin represents the solution set of the inequation 2x + 2y ≤ 80. Region represented by 2x + 4y ≤ 120: The line 2x + 4y = 120 meets the coordinate axes at C(60, 0) and D(0, 30) respectively. By joining these points we obtain the line 2x + 4y ≤ 120. Clearly (0, 0) satisfies the inequation 2x + 4y ≤ 120. So,the region containing the origin represents the solution set of the inequation 2x + 4y ≤ 120. The feasible region determined by the system of constraints, 2x + 2y ≤ 80, 2x + 4y ≤ 120 are as follows:
The corner points of the feasible region are O(0, 0), A(40, 0), E(20, 20) and D(0, 30). The values of Z at these corner points are as follows:
Corner point
Z = 3x +4y
O(0, 0)
3 × 0 + 4 × 0 = 0
A(40, 0)
3 × 40 + 4 × 0 = 120
E(20, 20)
3 × 20 + 4 × 20 = 140
D(0, 30)
10 × 0 + 4 × 30 = 120
We see that the maximum value of the objective function Z is 140 which is at E(20, 20) that means at x = 20 and y = 20. Thus, the optimal value of Z is 140.
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Question 395 Marks
A merchant plans to sell two types of personal computers a desktop model and a portable model that will cost Rs. 25,000 and Rs. 40,000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs. 70 lakhs and his profit on the desktop model is Rs. 4500 and on the portable model is Rs. 5000. Make an LPP and solve it graphically.
Answer
Let x and y be the number of desktop model and portable model respectively.
Number of desktop model and portable model cannot be negative.
Therefore,
x ≥ 0, y ≥ 0:
It is given that the monthly demand will not exist 250 units.
$\therefore$ x + y ≤ 250
Cost of desktop and portable model is Rs. 25,000 and Rs. 40,000 respectively.
Therefore, cost of x desktop model and y portable model  is Rs. 25,000 and Rs. 40,000 respectively and he does not want to invest more than Rs. 70 lakhs.
25000x + 40000y ≤ 7000000
Profit on the desktop model is Rs. 4500 and on the portable model is Rs. 5000.
Therefore, profit made by x desktop model and y portable model is Rs. 4500x and Rs. 5000y respectively.
Total profit = Z = 4500x + 5000y
The mathematical form of the given LPP is:
Maximize Z = 4500x + 5000y
Subject to constraints:
x + y ≤ 250
25000x + 40000y ≤ 7000000
x ≥ 0, y ≥ 0:
First we will convert inequations into equations as follows:
x + y = 250, 25000x + 40000y = 7000000, x = 0 and y = 0
Region represented by x + y ≤ 250:
The line x + y = 250 meets the coordinate axes at A(250, 0) and B(0, 250) respectively.
By joining these points we obtain the line x + y = 250.
Clearly (0, 0) satisfies the x + y = 250.
So, the region which contains the origin represents the solution set of the inequation x + y ≤ 250.
Region represented by 25000x + 40000y ≤ 7000000:
The line 25000x + 40000y = 7000000 meets the coordinate axes at C(280, 0) and D(0, 175) respectively.
By joining these points we obtain the line 25000x + 40000y = 7000000.
Clearly (0, 0) satisfies the inequation 25000x + 40000y ≤ 7000000.
So, the region which contains the origin represents the solution set of the inequation 25000x + 40000y ≤ 7000000.
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 ≤ 250, 25000x + 40000y ≤ 7000000, x ≥ 0 and y ≥ 0 are as follows.

The corner points are O(0, 0), D(0, 175), E(200, 50) and A(250, 0).
The values of the objective function Z at corner points of the feasible region are given in the following table:
Corner point
Z = 4500x + 5000y
 
O(0, 0)
0
 
D(0, 175)
875000
 
E(200, 50)
1150000
→ Maximum
A(250, 0)
1125000
  
Clearly, Z is maximum at x = 200 and y = 50 and the maximum value of Z at this point is 1150000.
Thus, 200 desktop models and 50 portable units should be sold to maximize the profit.
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Question 405 Marks
A firm manufactures two products, each of which must be processed through two departments, 1 and 2. The hourly requirements per unit for each product in each department, the weekly capacities in each department, selling price per unit, labour cost per unit, and raw material cost per unit are summarized as follows:
 
 
Product A
Product B
Weekly capacity
Department 1
3
2
130
Department 2
4
6
260
Selling price per unit
Rs. 25
Rs. 30
 
Labour cost per unit
Rs. 16
Rs. 20
 
Raw material cost per unit
Rs. 4
Rs. 4
 
The problem is to determine the number of units to produce each product so as to maximize total contribution to profit. Formulate this as a LPP.
Answer
Given  information can be tabulated below:
Product
Department 1
Department 2
Selling price
Labour cost
Row material cost
A
3
4
25
16
4
B
2
6
30
20
4
Capacity
130
260
 
 
  
Let the required product of product A and B be x and y units respectively.
Given, labour cost and raw material cost of one unit of product A is Rs 16 and Rs 4, so total cost of product A is Rs 16 + RS 4 = Rs 20 And given selling price of 1 unit of product A is Rs 25, So, profit on one unit of product.
A - 25 - 20 - Rs 5
Again, given labour cost and raw material cost of one unit of product B is Rs 20 and Rs 4 So, that cost of product B is Rs 20 + RS 4 - Rs 24 And given selling price of 1 unit of product B is Rs 30 So, profit on one unit of product B = 30 - 24 = Rs 6
Hence, profits on x unit of product A and y units of product B are Rs 5x and Rs 6y respectively.
Let z be the total profit, so Z = 5x + 6y
Given, production of one unit of product A and B need to process for 3 and 4 hours respectively in department 1, so production of x units of product A and y units of product 8 need to process for 3x and 4 hours respectively in Department 1. But to tal capacity of Department 1 is 130 hour, So, $\text{3x}+\text{2y}\leq130$  (First constraint)
Given, production of one unit of product A and B need to process for 4 and 6 hours respectively in department 2, so production of x units of product A and y units of product B need to process for 4x and 6y hours respectively in Department 2 but total capacity of Department 2 is 260 hours So, $\text{4x}+\text{6y}\leq 260$  (Second constraint)
Hence, mathematical formulation of LPP IS, Find x and y which Maximize z = 5x + 6y
Subject to constraint,
$\text{3x}+\text{2y}\leq130$
$\text{4x}+\text{6y}\leq260$
$\text{X,Y}\geq 0$ [Since production cannot be less than zero]
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Question 415 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 425 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 435 Marks
A medical company has factories at two places $, A$ and $B$. From these places, supply is made to each of its three agencies situated at $P, Q$ and $R$. The monthly requirements of the agencies are respectively $40,$ $40$ and $50$ packets of the medicines, while the production capacity of the factories $, A$ and $B,$ are $60$ and $70$ packets respectively. The transportation cost per packet from the factories to the agencies are given below:
How many packets from each factory be transported to each agency so that the cost of transportation is minimum? Also find the minimum cost?
Answer
The given information can be exhibited diagramatically as below:
Let factory $A$ transports $x$ packets to agency $P$ and $y$ packet to agency $Q$.
Since factory $A$ has capacity of $60$ packets so, rest $[60 - (x + y)]$ packets transported to agency $R$.
Since requirements are always non negative so,
$= x , y \geq 0 \ ($first constraint$)$
and
$60 = (x + y) \geq 0$
$(x + y) \geq 60 \ ($second constraint$)$
Since requirement of agency $P$ is $40$ packet but it has recieved $x$ packet, so $(40 - x)$ packets are transported from factory $B,$ requirement of agency $Q$ is $40$ packets but it has recieved $y$ packets, so $(40 - y)$ packets are transported from factory $B$.
Requirement of agency $R$ is $50$ packets but it has recieved $(60 - x - y)$ packets from factory $A,$ so $50 - (60 - x - y) - (x + y - 10)$ is transported from factory $B,$
As the requirements of agencies $P, Q, R$ are always non negative, so,
$40 - x \geq 0$
$X \leq 40 \ ($third constraint$)$
$40- y \geq 0$
$y \leq 40 \ ($fourth constraint$)$
$x + y - 10 \geq 0$
$x + y \geq 10 \ ($fifth constraint$)$
Costs of transportation of each packet from factory $A$ to agency $P, Q, R$ are $Rs. 5,4,3$ respectively and costs of transportation of each packet from factory $8$ to agency $P, Q, R$ are $Rs. 4, 2, 5$ respectively,
Let $Z$ be total cost of transportation so,
$Z = 5x + 4y + 3(60 - x - y) + 4(40 - x) + 2(40 - y) + 5(x + y - 10)$
$= 5x + 4y + 180 - 3x - 3y + 160 - 4x + 80 - 2y + 5x + 5y - 50$
$= 3x + 4y + 370$
Hence, mathematical formulation of $\text{LPP}$ is find $x$ and $y$ which
Maximize $Z = 3x + 4y + 370$
Subject to constraints,
$x, y \geq 0$
$x + y \leq 60$
$x \leq 40$
$y \leq 0$
$x + y \geq 10$
Region $ x, y \geq 0:$
It is represents first quandrant.
Region $x + y \leq 60:$
Line $x + y \leq 60$ meets axes at $A_1(60, 0), B_1(0, 60)$ respectively.
Region containing origin represents $x +y \leq 60$ as $(0, 0)$ satisfies $x + y \leq 60$
Region $X \leq 40$:
Line $x = 40$ is parallel to $y-$ axis and meets $x-$ axis at $A_2(40, 0)$.
Region containing origin represents $x \leq 40$ as $(0, 0)$ satisfies $ x \leq 40.$
Region $y \leq 40$:
line $y = 40$ is parallel to $x-$ axis and meets $y-$ axis at $B_2(0, 40)$.
Region containing origin represents $y \leq 40$ as $(0, 0)$ satisfies $y \leq 40$.
Region $x + y \geq 10$:
Line $x + y = 10$ meets axes at $A_2(10, 0), B_3(0, 10)$ respectively.
Region containing origin represents $x + y \geq 10$ as $(0, 0)$ does not satisfy $x + y \geq 10$.
Shaded region $A_2A_2PQB_2B_3$ represents feasible region.
Point $P(40, 20)$ is obtained by solving $x = 40$ and $x + y = 60$
Point $Q(20, 40)$ is obtained by solving $y = 40$ and $x + y = 60$

The value of $Z = 3x + 4y + 370$ at
$A_3(10, 0) = 3(10) + 4(0) + 370 = 400$
$A_2(40, 0) = 3(40) + 4(0) + 370 = 490$
$P(40, 20) = 3(40) + 4(20) + 370 = 570$
$Q(20, 40) = 3(20) + 4(40) + 370 = 590$
$B_2(0, 40) = 3(0) + 4(40) + 370 = 530$
$B_3(0, 10) = 3(0) + 4(10) + 370 = 410$
Minimum $Z = 400$ at $x = 10, y = 0$
From $A \rightarrow P = 10$ packets
From $A \rightarrow Q = 0$ packets
From $A \rightarrow R = 50$ packets
From $B \rightarrow P = 30$ packets
From $B \rightarrow Q = 40$ packets
From $B \rightarrow R = O$ packets
Minimum cost $= Rs. 400$
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Question 445 Marks
A manufacturer considers that men and women workers are equally efficient and so he pays them at the same rate. He has 30 and 17 units of workers (male and female) and capital respectively, which he uses to produce two types of goods A and B. To produce one unit of A, 2 workers and 3 units of capital are required while 3 workers and 1 unit of capital is required to produce one unit of B. If A and B are priced at Rs. 100 and Rs. 120 per unit respectively, how should he use his resources to maximise the total revenue? Form the above as an LPP and solve graphically. Do you agree with this view of the manufacturer that men and women workers are equally efficient and so should be paid at the same rate?
Answer
Let x units of A and y units of B be produced by the manufacturer.
The price of one unit of A is Rs. 100 and the price of one unit of B is Rs. 120.
Therefore, the total price of x units of A and y units of B or the total revenue is Rs. (100x + 120y).
One unit of A requires 2 workers and one unit of B requires 3 workers.
Therefore, x units of A and y units of B requires (2x + 3y) workers.
But, the manufacturer has 30 workers.
$\therefore$ 2x + 3y ≤ 30
Similarly, one unit of A requires 3 units of capital and one unit of B requires 1 unit of capital.
Therefore, x units of A and y units of B requires (3x + y) units of capital.
But, the manufacturer has 17 units of capital.
$\therefore$ 3x + y ≤ 17
Thus, the given linear programming problem is
Maximise Z = 100x + 120y
Subject to the constraints
2x + 3y ≤ 30
3x + y ≤ 17
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 O(0, 0), A(0, 10), B(173, 0) and C(3, 8).
The value of the objective function at these points are given in the following table:
Corner point
Z = 100x + 120y
(0, 0)
100 × 0 + 120 × 0 = 0
(0, 10)
100 × 0 + 120 × 10 = 1200
(173, 0)
100 × 173 + 120 × 0 = 17003
(3, 8)
100 × 3 + 120 × 8 = 1260
The maximum value of Z is 1260 at x = 3, y = 8.
Hence, the maximum total revenue is Rs. 1,260 when 3 units of A and 8 units of B are produced.
Yes, because the efficiency of a worker does not depend on whether the worker is a male or a female.
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Question 455 Marks
Solve the following LPP graphically:
Maximize Z = 20x + 10y
Subject to the following constraints
$\text{x}+2\text{y}\leq28$
$3\text{x}+\text{y}\leq24$
$\text{x}\geq2$
$\text{x},\text{y}\geq0$
Answer
The given contraints are:
$\text{x}+2\text{y}\leq28$
$3\text{x}+\text{y}\leq24$
$\text{x}\geq2$
$\text{x},\text{y}\geq0$
Converting the given inequation into equation, we get
x + 2y = 28, 3x + y = 24, x = 2 and y = 0
These lines are drawn on the graph and the shaded region ABCD represents the feasible region of the given LPP.

It can be observed that the feasible region is bounded.
The coordinates of the corner points of the feasible region are A(2, 13), B(2, 0), C(4, 12) and D(8, 0).
The values of the objective function, Z at these corner points are given in the following table:
Corner point Value of the objective Function Z = 20x + 10y
A(2, 13) Z = 20 × 2 + 10 × 13 = 170
B(2, 0) Z =20 × 2 + 10 × 0 = 40
C(8, 0) Z = 20 × 8 + 10 × 0 = 160
D(4, 12) Z = 20 × 4 + 10 × 12 = 200
From the table, Z is maximum at x = 4 and y = 12 and the maximum value of Z is 200.
Thus, the maximum value of Z is 200.
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Question 465 Marks
Maximum Z = 15x + 10y
Subject to
$3\text{x}+2\text{y}\leq80$
$2\text{x}+3\text{y}\leq70$
$\text{x},\text{y}\geq0$
Answer
First, we will convert the given inequations into equations, we obtain the following equations:
3x + 2y = 80, 2x + 3y = 70, x = 0 and y=0
Region represented by $3\text{x}+2\text{y}\leq80:$
The line 3x + 2y = 80 meets the coordinate axes at $\text{A}\Big(\frac{80}{3},0\Big)$ and B(0, 40) respectively.
By joining these points we obtain the line 3x + 2y = 80.
Clearly (0,0) satisfies the inequation $3\text{x}+2\text{y}\leq80$.
So, the region containing the origin represents the solution set of the inequation $3\text{x}+2\text{y}\leq80$.
Region represented by $2\text{x}+3\text{y}\leq70:$
The line 2x + 3y = 70 meets the coordinate axes at C(35, 0) and $\text{D}\Big(0,\frac{70}{3}\Big)$ respectively.
By joining these points we obtain the line $2\text{x}+3\text{y}\leq70$.
Clearly (0,0) satisfies the inequation $2\text{x}+3\text{y}\leq70$.
So, the region containing the origin represents the solution set of the inequation $2\text{x}+3\text{y}\leq70$.
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 $3\text{x}+2\text{y}\leq80$, $2\text{x}+3\text{y}\leq70$, $\text{x}\geq0$ and $\text{y}\geq0$ are as follows.

The corner points of the feasible are O(0, 0), $\text{A}\Big(\frac{80}{3},0\Big)\text{E}(20,10)$ and $\text{D}\Big(0,\frac{700}{3}\Big)$ .
The values of Z at these corner point are as follows.
$\text{Corner point}$
$\text{Z}=15\text{x}+10\text{y}$
$\text{O}(0, 0)$
$15\times0+10\times0=0$
$\text{A}\Big(\frac{80}{3},0\Big)$
$15\times\frac{80}{3}+10\times0=400$
$\text{E}(20, 10)$
$15\times20+10\times10=400$
$\text{D}\Big(0,\frac{70}{3}\Big)$
$15\times0+10\times\frac{70}{3}=\frac{700}{3}$
We see that maximum value of the objective functioin Z is 400 which is at $\text{A}\Big(\frac{80}{3},0\Big)$ and E(20, 10).
Thus, the optimal value of Z is 400.
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Question 475 Marks
Anil wants to invest at most Rs. 12000 in Saving Certificates and National Saving Bonds. According to rules, he has to invest at least Rs. 2000 in Saving Certificates and at least Rs. 4000 in National Saving Bonds. If the rate of interest on saving certificate is 8% per annum and the rate of interest on National Saving Bond is 10% per annum, how much money should he invest to earn maximum yearly income? Find also his maximum yearly income.
Answer
Let Anil invests Rs. x in Saving certificates and Rs. y in National Saving bonds.
Therefore, 
x, y ≥ 0
Anil wants to invest at most Rs. 12000 in Saving Certificates and National Saving Bonds.
x + y ≤ 12000
According to rules, he has to invest at least Rs. 2000 in Saving Certificates and at least Rs. 4000 in National Saving Bonds.
x ≥ 2000
y ≥ 4000
If the rate of interest on saving certificate is 8% per annum and the rate of interest on National Saving Bond is 10% per annum.
Total earning from investment $=\text{Z}=\frac{8\text{x}}{100}+\frac{10\text{y}}{100}$ which is to be maximised.
Thus, the mathematical formulat​ion of the given linear programmimg problem is 
Max $\text{Z}=\frac{8\text{x}}{100}+\frac{10\text{y}}{100}$ 
Subject
x + y ≤ 12000
x ≥ 2000
y ≥ 4000
First we will convert inequations into equations as follows:
x + y = 12000, x = 2000, y = 4000, x = 0 and y = 0
Region represented by x + y ≤ 12000:
The line x + y = 12000 meets the coordinate axes at A(12000, 0) and B(0, 12000) respectively.
By joining these points we obtain the line x + y = 12000.
Clearly (0, 0) satisfies the inequation x + y ≤ 12000.
So, the region which contains the origin represents the solution set of the inequation x + y ≤ 12000.
Region represented by x ≥ 2000:
The line x  = 2000 is the line that passes through (2000, 0) and is parallel to Y axis.
The region to the right of the line x = 2000 will satisfy the inequation x ≥ 2000.
Region represented by y ≥ 4000:
The line y = 4000 is the line that passes through (0, 4000) and is parallel to X axis.
The region above the line y = 4000 will satisfy the inequation y ≥ 4000.
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 is

The corner points are E(2000, 10000), D(8000, 4000), C(2000, 4000)
The values of Z at these corner points are as follows.
$\text{Corner point}$
$\text{Z}=\frac{8\text{x}}{100}+\frac{10\text{y}}{100}$
E
1160
D
1040
C
560
The maximum value of Z is 1160 which is attained at E(2000, 10000).
Thus, the maximum earning is Rs. 1160 obtained when Rs. 2000 were invested in Saving's certificates and Rs. 10000 were invested in National Saving Bond.
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Question 485 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 495 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 505 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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