Maths 4 Marks

Coverting the given inequation into equations, we get

2x + 3y = 13, 3x + y = 5 and x = 0, y = 0

Region represented by $2\text{x}+3\text{y}\leq13:$

The line meets coordinate axes at $\text{A}_1\Big(\frac{13}{2},0\Big)$ and $\text{B}_1\Big(0,\frac{13}{3}\Big)$ respectively.

Join these points to obtain the line 2x + 3y = 13, clearly, (0,0) satisfies the in eqation $2\text{x}+3\text{y}\leq13$, so, the region in xy-plane that contains origin represents the solution set of $2\text{x}+3\text{y}\leq13$.

Region represented by $3\text{x}+\text{y}\leq5:$

The line meets coordinate axes at $\text{A}_2\Big(\frac{5}{3},0\Big)$ and B₂(0, 5) respectively.

Join these points to obtain the line 3x + y = 5, clearly, (0, 0) satisfies the in eqation $3\text{x}+\text{y}\leq5$, so, the region in xy-plane that contains origin represents the solution set of $3\text{x}+\text{y}\leq5$.

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

It clearly represent first quadrant of xy-plane.

The common region to regions represented by above in equalities.

The coordinates of the corner points of the shaded region are $\text{O}(0,0),\text{A}\Big(\frac{5}{3},0\Big),\text{P}\Big(\frac{2}{7},\frac{29}{7}\Big),\text{B}_2\Big(0,\frac{13}{3}\Big)$.

The value of Z = 9x + 3y at

$\text{O}(0,0)=9(0)+3(0)=0$

$\text{A}_1\Big(\frac{5}{3},0\Big)=9\Big(\frac{5}{3}\Big)+3(0)=15$

$\text{P}\Big(\frac{2}{7},\frac{29}{7}\Big)=9\Big(\frac{2}{7}\Big)+3\Big(\frac{29}{7}\Big)=15$

$\text{B}_2\Big(0,\frac{13}{3}\Big)=9(0)+3\Big(\frac{13}{3}\Big)=13$

Clearly, Z is maximum at every point on the line joining A₁ and P, So

$\text{x}=\frac{2}{7}$ or $\frac{2}{7}$, $\text{y}=0$ or $\frac{29}{7}$

and maximum Z = 15.

Let required production of product A be x units and production of product B be y units.

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

Let Z be the total profit, so

Z = 60x + 40y

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

5x + 3y = 45000 (First constraint)

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

$\times\leq7000$ (Second constraint)

Hence, mathematical formulation of LPP is,

Find x and y which

maximize Z = 60x + 40y

Subject to constraints,

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

$\text{x}\leq7000$

$\text{y}\leq10000$

$\text{x},\text{y}\geq0$ [Since production can not be less than zero].

Let the factory manufacture x screws of type A and y screws of type B on each day.

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

The given information can be compiled in a table as follows.

The profit on a package of screws A is Rs. 7 and on the package of screws B is Rs. 10.

Therefore, the constraints are

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

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

Total profit, Z = 7x + 10y.

The mathematical formulation of the given problem is,

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

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

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

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

The feasible region determined by the system of constraints is

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

The values of Z at these corner points are as follows.

The maximum value of Z is 410 at (30, 20).

Thus, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs 410.

Let x units of product A and y units of product B were manufactured.

Number of units cannot be negative.

Therefore,  x, y ≥ 0

According to question, the given information can be tabulated as

Also, the availability of time is 40 hours and the revenue should be atleast Rs 10000.

Further, it is given that there is a permanent order for 14 units of product A and 16 units of product B.

Therefore, the constraints are

$200\text{x}+300\text{y}\geq10000$

$0.5\text{x}+\text{y}\leq40$

$\text{x}\geq14$

$\text{y}\geq16$

If the profit on each of product A is Rs 20 and on product B is Rs 30.Therefore, profit gained on x units of product A and y units of product B is Rs 20x and Rs 30y respectively.

Total profit = Z = 20x+30y which is to be maximised

Thus, the mathematical formulat​ion of the given linear programmimg problem is 

Max Z = 20x + 30y

Subject to

$2\text{x}+3\text{y}\geq100$

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

$\text{x}\geq14$

$\text{y}\geq16$

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

First we will convert inequations into equations as follows:

2x + 3y = 100

x + 2y = 80

x = 14

y = 16

x = 0

y = 0

Region represented by 2x + 3y ≥ 100:

The line 2x + 3y = 100 meets the coordinate axes at A₁(50, 0) and $\text{B}_1\Big(0,\frac{100}3{}\Big)$ respectively.

By joining these points we obtain the line 2x + 3y = 100.

Clearly (0, 0) does not satisfies the 2x + 3y = 100.

So, the region which does not contain the origin represents the solution set of the inequation 2x + 3y ≥ 100.

Region represented by x + 2y ≤ 80:

The line x + 2y = 80 meets the coordinate axes at C₁(80, 0) and D₁(0, 40) respectively.

By joining these points we obtain the line x + 2y = 80.

Clearly (0, 0) satisfies the inequation x + 2y ≤ 80.

So, the region which contains the origin represents the solution set of the inequation x + 2y ≤ 80.

Region represented by x ≥ 14

x = 14 is the line passes through (14, 0) and is parallel to the Y axis.

The region to the right of the line x = 14 will satisfy the inequation.

Region represented by y ≥ 16

y = 16 is the line passes through (0, 16) and is parallel to the X axis.

The region above the line y = 16 will satisfy the inequation.

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 + 3y ≥ 100, x + 2y ≤ 80, x ≥ 14, y ≥ 16, x ≥ 0 and y ≥ 0 are as follows.

The corner points of the feasible region are E₁(26, 16), F₁(48, 16), G₁(14, 33) and H₁(14, 24)

The values of Z at these corner points are as follows.

The maximum value of Z is Rs 1440 which is attained at F₁(48, 16).

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.

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.

Let the company manufacture x souvenirs of type A and y souvenirs of type B.

Therefore, $\text{x}\geq0$ and $\text{y}\geq0$ The given information can be complied in a table as follows.

The profit on type A souvenirs is Rs. 5 and on type B souvenirs is Rs. 6.

Therefore, the constraints are

$5\text{x}+8\text{y}\leq200$

$10\text{x}+8\text{y}\leq240$

i.e., $5\text{x}+4\text{y}\leq120$

Total profit, Z = 5x + 6y

The mathematical formulation of the given problem is

Maximize Z = 5x + 6y ... (1)

Subject to the constraints,

$5\text{x}+8\text{y}\leq200 \dots (2)$

$5\text{x}+4\text{y}\leq120\dots(3)$

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

The feasible region determined by the system of constraints is as follows.

The corner point are A(24, 0), B(8, 20), and C(0, 25).

The values of Z at these corner points are as follows.

The maximum value of Z is 200 at (8, 20).

Thus, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day to get the maximum profit of Rs. 160.

Let x trunks of first type and y trunks of second type were manufactured.

Number of trunks cannot be negative.

Therefore,

x, y ≥ 0

According to question, the given information can be tabulated as

Therefore, the constraints are

$3\text{x}+3\text{y}\leq18$

$3\text{x}+2\text{y}\leq15$

He earns a profit of Rs. 30 and Rs. 25 per trunk of the first type and the second type respectively.

Therefore, profit gained by him from x trunks of first type and y trunks of second type is Rs. 30x and Rs. 25y respectively.

Total profit = Z = 30x + 25y which is to be maximised

Thus, the mathematical formulat​ion of the given linear programmimg problem is

Max Z = 30x + 25y

Subject to

$3\text{x}+3\text{y}\leq18$

$3\text{x}+2\text{y}\leq15$

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

First we will convert inequations into equations as follows:

3x + 3y = 18, 3x + 2y = 15, x = 0 and y = 0

Region represented by 3x + 3y ≤ 18:

The line 3x + 3y = 18 meets the coordinate axes at A₁(6, 0) and B₁(0, 6) respectively.

By joining these points we obtain the line 3x + 3y = 18.

Clearly (0, 0) satisfies the 3x + 3y = 18.

So, the region which contains the origin represents the solution set of the inequation 3x + 3y ≤ 18.

Region represented by 3x + 2y ≤ 15:
The line 3x + 2y = 15 meets the coordinate axes at C₁(5, 0) and $\text{D}_1\Big(0,\frac{15}{2}\Big)$ respectively.

By joining these points we obtain the line 3x + 2y = 15.

Clearly (0, 0) satisfies the inequation 3x + 2y ≤ 15.

So, the region which contains the origin represents the solution set of the inequation 3x + 2y ≤ 15.

Region represented by x ≥ 0 and y ≥ 0:

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

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

The feasible region determined by the system of constraints 3x + 3y ≤ 18, 3x + 2y ≤ 15, x ≥ 0 and y ≥ 0 are as follows.

The corner points are O(0, 0), B₁(0, 6), E₁(3, 3) and C₁(5, 0).

The values of Z at these corner points are as follows.

Let 25x grams of food F₁ and 25y grams of food F₂ be used to fulfil the minimum requirement of thiamine, phosphorus and iron.

As, we are given,

And the minimum requirement of the nutrients in the diet are 1.00mg of thiamine, 7.50mg of phosphorous and 10.00mg of iron.

Therefore, $0.25\text{x}+0.10\text{y}\geq1$

$0.75\text{x}+1.50\text{y}\geq7.5$

$1.6\text{x}+0.8\text{y}\geq10$

Since the quantity cannot be negative

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

The cost of F₁ is 20 paise per 25gms while the cost of F₂ is 15 paise per 25 gms.

Therefore, the cost of 25x grams of food F₁ and 25y grams of food F₂ is Rs. (0.20x + 0.15y).

Hence,

Minimize Z = 0.20x + 0.15y

Subject to

$0.25\text{x}+0.10\text{y}\geq1,0.75\text{x}+1.50\text{y}\geq7.5,$ $1.6\text{x}+0.8\text{y}\geq10,\text{x},\text{y}\geq0.$

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

0.25x + 0.10y = 1, 0.75x + 1.50y = 7.5, 1.6x + 0.8y = 10, x = 0 and y = 0.

The line 0.25x + 0.10y = 1 meets the coordinate axis at A(4, 0) and B(0, 10).

Join these points to obtain the line 0.25x + 0.10y = 1.

Clearly, (0, 0) does not satisfies the inequation $0.25\text{x}+0.10\text{y}\geq1$.

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

The line 0.75x + 1.50y = 7.5. meets the coordinate axis at C(10, 0) and D(0, 5).

Join these points to obtain the line 0.75x + 1.50y = 7.5.

Clearly, (0, 0) does not satisfies the inequation $0.75\text{x}+1.50\text{y}\geq7.5$.

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

The line 1.6x + 0.8y = 10 meets the coordinate axis at $\text{E}\Big(\frac{25}{4},0\Big)$ and $\text{F}\Big(0,\frac{25}{2}\Big).$

Join these points to obtain the line 1.6x + 0.8y = 10.

Clearly, (0, 0) does not satisfies the inequation $1.6\text{x}+0.8\text{y}\geq10$.

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

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

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

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

These lines are drawn using a suitable scale.

The corner points of the feasible region are F(0, 12.5), G(5, 2.5), C(10, 0)

The value of the objective function at these points are given by the following table:

Thus, the minimum cost is at G which is Rs. 1.375.

Let required quantity of bran and rice be x kg and y kg.

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

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

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

Z = 5x + 4y

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

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

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

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

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

Hence, mathematical formulation of LPP is,

Find x and y which minimize

Z = 5x + 4y

subject to constraints,

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

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

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

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

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

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

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

The value of Z = 5x + 4y at

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

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

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

Smallest value of Z is 4.6.

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

Hence,

Minimum cost of mixture = Rs 4.6

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.

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.

Let the dietician wishes to mix x units of food F₁ and y kg of F₂.

Clearly, $\text{x},\text{y}\geq0$ 

The given information can be tabulated as follows:

The constraints are

$3\text{x}+6\text{y}\geq80$

$4\text{x}+3\text{y}\geq100$

It is given that cost of food F₁ and F₂ is Rs. 4 and Rs. 6 per unit respectively.

Therefore, cost of x units of food F₁ and y units of food F₂ is Rs. 4x and Rs. 6y respectively.

Let Z denote the total cost

$\therefore$ Z = 4x + 6y

Thus, the mathematical formulat​ion of the given linear programmimg problem is Minimize Z = 4x+6y

subject to

$3\text{x}+6\text{y}\geq80$

$4\text{x}+3\text{y}\geq100$

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

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

3x + 6y = 80, 4x + 3y = 100, x = 0 and y = 0

The line 3x + 6y = 80 meets the coordinate axis at $\text{A}\Big(\frac{80}{3},0\Big)$ and $\text{B}\Big(0,\frac{40}{3}\Big)$.

Join these points to obtain the line 3x + 6y= 80.

Clearly, (0, 0) does not satisfies the inequation $3\text{x}+6\text{y}\geq80$.

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

The line 4x + 3y = 100 meets the coordinate axis at C(25, 0) and $\text{D}\Big(0,\frac{100}{3}\Big)$.

Join these points to obtain the line 4x + 3y = 100.

Clearly, (0, 0) does not satisfies the inequation $4\text{x}+3\text{y}\geq100$.

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

Region represented by x ≥ 0 and y ≥ 0:

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

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

These lines are drawn using a suitable scale.

The feasible region determined by the system of constraints is

The corner points are $\text{D}\Big(0,\frac{100}{3}\Big),\text{E}\Big(24,\frac{4}{3}\Big)$ and $\text{A}\Big(\frac{80}{3},0\Big)$.

The values of Z at these corner points are as follows:

The minimum value of Z is Rs. 104 which is at $\text{E}\Big(24,\frac{4}{3}\Big)$.

Let x be the amount of food X and Y be the amount of food Y that is to be mixed which will produce the required diet.

Then the mathematical modal of the LPP is as follows:

Minimise Z = 16x + 20y

Subject to 

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

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

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

To solve the LPP we draw the lines,

x + 2y = 10

2x + 2y = 12

3x + y = 8

The feasible region of the LPP is shaded in graph.

The coordinates of the vertices (corner points) of the feasible region ABCD are A(10, 0), B(2, 4), C(1, 5) and D(0, 8).

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

2kg of food X and 4kg of food Y will be required to minimize the cost of the diet.

The least cost of the mixture is Rs. 112.

Let number of goods A and B are x and y respectively.

Since, profits on each A and B are Rs. 40 and Rs. 50 respectively.

So, profits on x of type A and y of type B are 40x and 50y respectively, Let Z be total profit on A and B, so,

Z = 40x + 50y

Since, each A and B require 3gm and 1gm of silver respectively.

So, x of type A and y type B require 3x and y gm silver respectively but, Total silver available is 9 gm. so,

$3\text{x}+\text{y}\leq9$ (first constraint)

Since, each A and B require 1gm and 2gm of gold respectively.

So, x of type A and y type B require x and 2y gm of gold respectively but, Total gold available is 8 gm, so,

$\text{x}+2\text{y}\leq8$ (second constraint)

Hence, mathematical formulation of LPP is,

Find x and y which maximize

Z = 40x + 50y

Subject to constraints,

$3\text{x}+\text{y}\leq9$

$\text{x}+2\text{y}\leq8$

$\text{x},\text{y}\geq0$ [Since production of A and B can not be less than zero]

Region $3\text{x}+\text{y}\leq9$:

Line 3x +y = 9 meets axes at A(3, 0), B(0, 9) respectively.

Region containing origin represents $3\text{x}+\text{y}\leq9$ as (0, 0) satisfies $3\text{x}+\text{y}\leq9$.

Region $\text{x}+2\text{y}\leq8$:

Line x +2y = 8 meets axes at A₂(8, 0), B₂(0, 4) respectively.

Region containing origin represents $\text{x}+2\text{y}\leq8$ as (0, 0) satisfies $\text{x}+2\text{y}\leq8$.

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

It represents first quadrant.

Shaded region OA₂PB₂ is the feasible region.

Point P(2, 3) is obtained by solving 3x + y - 9 and x + 2y = 8

The value of Z = 40x + 50y at

O(0, 0) = 40(0) + 50(0) = 0

A₁(3, 0) = 40(3) + 50(0) = 120

P(2, 3) = 40(2) + 50(3) = 230

B₂(0, 4) = 40(0) + 50(4) = 200

Therefore maximum Z = 230 at x = 2, y = 3

Hence,

Maximum profit = Rs. 230 number of goods of type A = 2, type B = 3

Let x articles of deluxe model and y articles of an ordinary model be made.

Number of articles cannot be negative.

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

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

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

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

The making of a deluxe model requires 2 hrs.

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

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

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

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

Max Z = 15x + 10y

Subject to

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

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

$\text{x}\geq0$

$\text{y}\geq0$

The feasible region determined by the system of constraints is:

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

The values of Z at these corner points are as follows.

The maximum value of Z is 300 which is attained at C(20, 0).

Thus, the maximum profit is Rs. 300 obtained when 10 units of deluxe modal and 20 unit of ordinary model is produced.

Let required production of product A, B and C bex,y and z units respectively.

Given, profit on one unit of product A, B and C are Rs. 3, RS. 2, RS. 4, so Profit on x unit of A, y unit of B and z unit of C are given by Rs. 3x, Rs. 2y, Rs. 4z.

Let U be the total profit, so

U = 3x + 2y + 4z

Given, one unit of product A, B and C requires 4,3 and 5 minutes on machine M. So, x units of product A, y units of B and z units of product C need 4x, 3y and 5z minutes on machine M, is 2000 minutes, so

$4\text{x}+3\text{y}+5\text{z}\leq200$ (First constraint)

Given, one unit of product A, B and C requires 2,2 and 4 minutes on machine M2. So, X units of A, y units of B and z units of C require 2x, 2y and 4z minutes on machine M2 is 2500 minutes, so

$2\text{x}+2\text{y}+4\text{z}\leq2500$ (Second constraint)

Also, given that firm must manufacture 100 A's, 200 B's and 50 C's but not more than 150 A's.

$100\leq\times\leq150$

$\text{y}\geq200$ (Other constraints)

$\text{z}\geq50$

Hence, mathematical formulation of LPP is:-

Find x, y and z which

maximize U = 3x + 2y + 4z

Subject to constraints,

$4\text{x}+3\text{y}+5\text{z}\leq2000$

$2\text{x}+2\text{y}+4\text{z}\leq2500$

$100\leq\times\leq150$

$\text{y}\geq200$

$\text{z}\geq50$

And, $\text{x},\text{y},\text{z}\geq0$ [Since, x, y, z are non-negative]

Let required quantity of compound A and B are x and y kg.

Since, cost of one kg of compound A and B are Rs. 4 and Rs. 6 per kg.

So, cost of x kg. of compound A and y kg. of compound B are Rs. 4x and Rs by respectively,

Let Z be the total cost of compounds, so,

Z = 4x + 6y

Since, compound A and B contain 1 and 2 units of ingredient c per kg, respectively, so, x kg. of compound A and y kg. of compound B contain x and 2y units of ingredient C respectively but minimum requirement of ingredient C is 80 units, so,

$\text{x}+2\text{y}\geq80$ (first constraint)

Since, compound A and B contain 3 and 1 unit of ingredient D per kg, respectively, so, x kg. of compound A and y kg. of compound B contain 3x and y units of ingredient respectively but minimum requirement of ingredient D is 75 units, so,

$3\text{x}+\text{y}\geq75$ (second constraint)

Hence, mathematical formulation of LPP is,

Find x and y which minimize

Z = 4x + 6y

Subject to constraints,

$\text{x}+2\text{y}\geq80$

$3\text{x}+\text{y}\geq75$

$\text{x},\text{y}\geq0$ [Since production can not be less than zero]

Region $\text{x}+2\text{y}\geq80$:

Line x + 2y = 80 meets axes at A₁(80, 0), B₁(0, 40) respectively.

Region not containing origin represents $\text{x}+2\text{y}\geq80$ as (0, 0) does not satisfy $\text{x}+2\text{y}\geq80$.

Region $3\text{x}+\text{y}\geq75$:

Line 3x +y = 75 meets axes at A₂(25, 0), B₂(0, 75) respectively.

Region not containing origin represents $3\text{x}+\text{y}\geq75$ as (0, 0) does not satisfy $3\text{x}+\text{y}\geq75$.

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

It represents first quadrant.

Unbouded shaded region A₁PB₂ represents feasible region point is obtained by solving x + 2y = 80 and 3x + y = 75

The value of Z = 4x + 6y at

A₁(80, 0) = 4(80) + 6(0) = 320

P(14, 33) = 4(14) + 6(33) = 254

B₂(0, 75) = 4(0) + 6(75) = 450

Smallest value of Z = 254 open half plane 4x + 6y < 254 has no point in common with feasible region.

So,

Smallest value is the minimum value.

Minimum cost = Rs. 254

Quantity of A = 14kg

Quantity of B = 33kg.

Let tailor A work for x days and tailor B work for y days.

In one day, A can stitch 6 shirts and 4 pants whereas B can stitch 10 shirts and 4 pants. Thus, in x days A can stitch 6x shirts and 4x pants. Similarly, in y days B can stitch 10y shirts and 4y pants.

It is given that the minimum requirement of the shirts and pants are respectively 60 and 32 respectively.

Thus,

$6\text{x}+10\text{y}\geq60,4\text{x}+4\text{y}\geq32$

Further it is given that A and B earn Rs 150 and Rs 200 per day respectively. Thus, in x days and y days, A and B earn Rs 150x and Rs 200y respectively.

Let Z denotes the total cost

$\therefore$ Z = Rs 150x + 200 y

Number of days cannot be negative.

Therefore, $\text{x},\text{y}\geq0$

Hence, the required LPP is as follows:

Minimize Z=150x + 200 y

Subject to

$6\text{x}+10\text{y}\geq60$

$4\text{x}+4\text{y}\geq32$

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

Let the person takes x lbs and y lbs of food I and II respectively that were taken in the diet. Since, per lb of food I costs Rs 60 and that of food II costs Rs 100.

Therefore, x lbs of food I costs Rs 60x and y lbs of food II costs Rs 100y.

Total cost per day = Rs (60x + 100y)

Let Z denote the total cost per day

Then, Z = 60x + 100 y

Total amount of calcium in the diet is 10x+5y

Since, each lb of food I contains 10 units of calcium. Therefore, x lbs of food I contains 10x units of calciumn.

Each lb of food II contains 5 units of calciu.So,y lbs of food II contains 5y units of calcium.

Thus, x lbs of food I and y lbs of food II contains 10x + 5y units of calcium.

But, the minimum requirement is 20 lbs of calcium.

$\therefore10\text{x}+5\text{y}\geq20$

Since, each lb of food I contains 5 units of protein. Therefore, x lbs of food I contains 5x units of protein,

Each lb of food II contains 4 units of protein. So,y lbs of food II contains 4y units of protein,

Thus, x lbs of food I and y lbs of food II contains 5x + 4y units of protein.

But, the minimum requirement is 20 lbs of protein.

$\therefore5\text{x}+4\text{y}\geq20$

Since, each lb of food I contains 2 units of calories. Therefore, x lbs of food I contains 2x units of calories.

Each lb of food II contains units of calories. So,y lbs of food II contains by units of calories.

Thus, x lbs of food I and y lbs of food II contains 2x + 6y units of calories.

But, the minimum requirement is 13 lbs of calories.

$\therefore2\text{x}+6\text{y}\geq13$

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 follows:

Min Z = 60x + 100y

subject to

$10\text{x}+5\text{y}\geq20$

$5\text{x}+4\text{y}\geq20$

$2\text{x}+6\text{y}\geq13$

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

Let x bags of fertilizer P and Y bags of fertilizer Q used in the garden to minimize the usage of nitrogen.

Then the mathematical modal of the LPP is as follows:

Minimise Z = 3x + 3.5y

Subject to 

$\text{x}+2\text{y}\geq240$

$3\text{x}+1.5\text{y}\geq270$

$1.5\text{x}+2\text{y}\leq310$

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

To solve the LPP we draw the lines,

x + 2y = 240

3x + 1.5y = 270

1.5x + 2y = 310

The feasible region of the LPP is shaded in graph.

The coordinates of the vertices (corner points) of the feasible region ABC are A(40, 100), B(140, 50), and C(20, 140).

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

40 bags of brand P and 100 bags of brand Q should be used to minimize.

The amount of nitrogen added to the garden.

The minimum amount of notrogen added in the garden is 470kg.

Let the dietician wishes to mix x kg of food X and y kg of Y.

Therefore, 

As we are given,

It is given that the mixture should contain at least 6 units of vitamin A, 7 units of vitamin B, 11 units of vitamin C and 9 units of vitamin D.

Therefore, the constrants are

$\text{x}+2\text{y}\geq6$

$\text{x}+\text{y}\geq7$

$\text{x}+3\text{y}\geq11$

$2\text{x}+\text{y}\geq9$

It is given that cost of food X is Rs 5 per kg and cost of food Y is Rs. 8 per kg.

Thus, Z= 5x+8y

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

Minimize Z= 5x+8y

subject to

$\text{x}+2\text{y}\geq6$

$\text{x}+\text{y}\geq7$

$\text{x}+3\text{y}\geq11$

$2\text{x}+\text{y}\geq9$

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

x + 2y = 6, x + y = 7, x + 3y = 11, 2x + y =9, x = 0 and y=0.

The line x + 2y = 6 meets the coordinate axis at A₁(6, 0) and B₁(0, 3).

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

Clearly, (0, 0) does not satisfies the inequation $\text{x}+2\text{y}\geq6$.

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

The line x + y = 7 meets the coordinate axis at C₁(7, 0) and D₁(0, 7).

Join these points to obtain the line x + y = 7.

Clearly, (0, 0) does not satisfies the inequation $\text{x}+\text{y}\geq7$.

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

The line x + 3y = 11 meets the coordinate axis at E₁(11, 0) and $\text{F}_1\Big(0,\frac{11}{3}\Big)$.

Join these points to obtain the line x + 3y = 11.

Clearly, (0, 0) does not satisfies the inequation $\text{x}+3\text{y}\geq11$.

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

The line 2x + y = 9 meets the coordinate axis at $\text{G}_1\Big(\frac{9}{2},0\Big)$ and H(0, 9).

Join these points to obtain the line 2x + y = 9.

Clearly, (0, 0) does not satisfies the inequation $2\text{x}+\text{y}\geq9$.

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

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

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

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

These lines are drawn using a suitable scale.

The feasible region determined by the system of constraints is

The corner points are H₁(0, 9), I₁(2, 5), J₁(5, 2), E₁(11, 0).

The values of Z at these corner points are as follows

The minimum value of Z is at J₁(5, 2) which is Rs. 41.

Hence, cheapest combination of foods will be 5 units of food X and 2 units of food Y.

Let x packets of food P and y packets of food Q be used to make the diet.

Each packet of food P contains 6 units of vitamin A and each packet of food Q contains 3 units of vitamin A.

Therefore, x packets of food P and y packets of food Q contains (6x + 3y) units of vitamin A.

Since each packet of food P contains 12 units of calcium and each packet of food Q contains 3 units of calcium, therefore, x packets food P and y packets of food Q will contain (12x + 4y) units of calcium.

But, the diet should contain atleast 240 units of calcium.

$\therefore12\text{x}+3\text{y}\geq240$

$\Rightarrow4\text{x}+\text{y}\geq80$

Similarly, x packets of food P and y packets of food Q will contain (4x + 207) units of iron. But, the diet should contain atleast 460 units of iron.

$\therefore4\text{x}+20\text{y}\geq460$

$\Rightarrow\text{x}+5\text{y}\geq115$

Also, x packets of food P and y packets of food Q will contain (6x + 4y) units of cholesterol.

But, the diet should contain atmost 300 units of cholesterol.

$\therefore6\text{x}+4\text{y}\leq300$

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

Thus, the given linear programming problem is

Minimise Z = 6x + 3y

subject to the constraints

$4\text{x}+\text{y}\geq80$

$\text{x}+5\text{y}\geq115$

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

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

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(2, 72), B(15, 20) and C(40, 15).

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

The smallest value of Z is 150 which is obtained at x = 15 and y = 20.

It can be verified that the open half-plane represented by $6\text{x}+3\text{y}\leq150$ has no common points with the feasible region.

Thus, 15 packets of food P and 20 packets of food Q should be used to minimise the amount of vitamin A in the diet.

Hence, the minimum amount of vitamin A is 150 units.

F₁ consists of 10% nitrogen and F₂ consists of 5% nitrogen.

But, the farmer needs atleast 14kg of nitorgen for the crops.

10% of x kg + 5% of y kg 14kg

$\Rightarrow\frac{\text{x}}{10}+\frac{\text{y}}{20}\geq14$

$\Rightarrow2\text{x}+\text{y}\geq280$

Similarly, F₁ consists of 6% phosphoric acid and F₂ consists of 10% phosphoric acid.

But, the farmer needs atleast 14kg 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 F₁ is Rs. 6/kg and fertilizer F₂ is Rs. 5/kg,

Therefore, total cost of x kg of fertilizer F₁ and and y kg of fertilizer F₂ is Rs. (6x + 5y).

Thus, the given linear programming problem is

Minimise Z = 6x + 5y

Subject to the constraints

2x + y ≥ 280

3x + 5y ≥ 700

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 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.

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, 100kg of fertilizer F₁ and 80kg of fretilizer F₂ should be used so that the nutrient requirements are met at minimum cost.

The minimum cost is Rs. 1,000.

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

3x + 5y = 15, 5x + 2 y = 10, x = 0 and y = 0

Region represented by $3\text{x}+5\text{y}\leq15:$

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

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

Clearly (0, 0) satisfies the inequation $3\text{x}+5\text{y}\leq15$.

So, the region containing the origin represents the solution set of the inequation $3\text{x}+5\text{y}\leq15$.

Region represented by $5\text{x}+2\text{y}\leq10:$

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

By joining these points we obtain the line 5x + 2y = 10.

Clearly (0, 0) satisfies the inequation $5\text{x}+2\text{y}\leq10$.

So, the region containing the origin represents the solution set of the inequation $5\text{x}+2\text{y}\leq10$.

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, $3\text{x}+5\text{y}\leq15,5\text{x}+2\text{y}\leq10,\text{x}\geq0$ and $\text{y}\geq0$,are as follows.

Therefore, he has to spend Rs. 120 at most on petrol.

$\therefore2\text{x}+3\text{y}\leq120\ ......(\text{i})$

Also, he has at most 1 hour time.

$\therefore\frac{\text{x}}{50}+\frac{\text{y}}{80}\leq1$

$\Rightarrow8\text{x}+5\text{y}\leq400\ .....(\text{ii})$

Also, we have $​​\text{x}\geq0,​​\text{y}\geq0$ [non-negative constraints]

Thus, required LPP to travel maximum distance by him is

 Maximise $\text{Z}=\text{x}+\text{y},$ subject to $2\text{x}+3\text{y}\leq120,8\text{x}+5\text{y}\leq400,\text{x}\geq0,\text{y}\geq0.$

Let the number of necklaces manufacture be x, and the number of bracelets manufacture be y.

Since the total number of items are at most 24.

$\text{x}+\text{y}\leq24\ ....(1)$

Bracelets takes 1 hour to manufacture and necklaces takes half an hour to manufacture.

x item takes x hour to manufacture and y items take y/2 hour to manufacture and maximum time available is 16 hours. therefore

$\text{x}_2+\text{y}\leq16\ ....(2)$

the profit on one necklace is Rs. 100 and the profit on one bracelet is Rs. 300

Let the profit be Z.

Now we wish to maximize the profit.

So,

Max Z = 100x + 300y ....(3)

So,

$\text{x}+\text{y}\leq24$

$\text{x}_2+\text{y}\leq16$

Max Z = 100x + 300 is the required L.P.P.

We have to maximize Z = 7x + 10y

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

x + y = 30000.y = 12000, x = 6000, x = y, x = 0 and y = 0.

Region represented by $\text{x}+\text{y}\leq30000$:

The line x + y = 30000 meets the coordinate axes at A(30000, 0) and B(0, 30000) respectively.

By joining these points we obtain the line x + y = 30000. Clearly (0, 0) satisfies the inequation $\text{x}+\text{y}\leq30000$.

So, the region containing the origin represents the solution set of the inequation $\text{x}+\text{y}\leq30000$.

The line y = 12000 is the line that passes through C(0, 12000) and parallel to x axis.

The line x = 6000 is the line that passes through (6000, 0) and parallel to y axis.

Region represented by $\text{x}\geq\text{y}$

The line x = y is the line that passes through origin.

The points to the right of the line x=y satisfy the inequation $\text{x}\geq\text{y}$.

Like by taking the point (-12000, 6000).

Here, 6000 > -12000 which implies y > x.

Hence, the points to the left of the line x = y will not satisfy the given inequation $\text{x}\geq\text{y}$.

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, $\text{x}+\text{y}\leq30000,\text{y}\leq12000,\text{x}\geq6000,\text{x}\geq\text{y},\text{x}\geq0$  and $\text{y}\geq0$ are as follows:

The corner points of the feasible region are D(6000, 0), A(3000,0), F(18000, 12000) and E(12000, 12000).

The values of Z at these corner points are as follows:

Let the number of cakes of one kind and another kind be x and y, respectively.

Therefore, the total number of cakes produced are (x + y).

One kind of cake requires 200g of flour and another kind of cake requires 100g of flour.

So, x cakes of one kind and ycakes of another kind requires (200x + 100y)g of flour.

But, cakes should contain atmost 5kg of flour.

$200\text{x}+100\text{y}\leq1000$

$2\text{x}+\text{y}\leq50$

One kind of cake requires 25g of fat and another kind of cake requires 50g of fat.

So, X cakes of one kind and y cakes of another kind requires (25x + 50y)g of fat.

But, cakes should contain atmost 1 kg of fat.

$25\text{x}+50\text{y}\leq1000$

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

Thus, the given linear programming problem is Minimise

Z= x + y

Subject to the constraints

$2\text{x}+\text{y}\leq50$

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

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

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

The coordinates of the corner points of the feasible region are O(0, 0), A(25, 0), B(20, 10) and C(0, 20).

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

Thus, the maximum value of Z is 30 at x = 20, y = 10.

Hence, the maximum number of cakes which can be made are 30.

Let x kg of type 1 fertilizer and y kg of type II fertilizer are supplied.

Quantity of fertilizer cannot be negative.

Therefore, x, y ≥ 0

A gardener has supply of fertilizer of type I which consists of 10% nitrogen and type II fertilizer consists of 5% nitrogen and he needs at least 14kg of nitrogen for his crop.

10 × 100 + 5 × 100 ≥ 14 = 10x + 5x ≥ 1400

A gardener has supply of fertilizer of type I which consists 6% phosphoric acid and type II fertilizer consists of 10% phosphoric acid.

And he needs 14kg of phosphoric acid for his crop.

6 × 100 + 10 × 100 ≥ 14 = 6x + 10x ≥ 1400

Therefore, according to the question, constraints are 10x + 5y ≥ 1400, 6x + 10y ≥ 1400

If the type I fertilizer costs 60 paise per kg and type II fertilizer costs 40 paise per kg.

Therefore, cost of x kg of type 1 fertilizer and y kg of type II fertilizer is Rs. 0.60x and Rs. 0.40y respectively.

Total cost = Z = 0.60x + 0.40y which is to be minimised.

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

Min Z = 0.60x + 0.40y

Subject to

6x + 10y ≥ 1400

10x + 5y ≥ 1400

x, y ≥ 0

First we will convert inequations into equations as follows:

6x + 10y = 1400, 10x + 5y = 1400, x = 0 and y = 0

Region represented by 6x +10y ≥ 1400:

The line 6x +10y = 1400 meets the coordinate axes at A(7003, 0) and B(0, 140) respectively.

By joining these points we obtain the line 6x + 10y = 1400.

Clearly (0, 0) does not satisfies the 6x + 10y = 1400.

So, the region which does not contain the origin represents the solution set of the inequation 6x + 10y ≥ 1400.

Region represented by 10x + 5y ≥ 1400:

The line 10x + 5y = 1400 meets the coordinate axes at C(140, 0) and D(0, 280) respectively.

By joining these points we obtain the line 10x + 5y = 1400.

Clearly (0, 0) does not satisfies the inequation 10x + 5y 1400.

So, the region which does not contain the origin represents the solution set of the inequation 10x + 5y ≥ 1400.

Region represented by x ≥ 0 and y ≥ 20:

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

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

The feasible region determined by the system of constraints 6x + 10y ≥ 1400, 10x + 5y ≥ 1400, X ≥ 0 and y ≥ 0 are as follows.

The corner points are D(0, 280), E(100, 80) and $\text{A}\Big(\frac{700}{3}, 0\Big).$

The values of Z at these corner points are as follows:

The minimum value of Z is Rs. 92 which is attained at E(100, 80).

Thus, the minimum cost is Rs. 92 obtained when 100kg of type I fertilizer and 80kg of type II fertilizer were supplied.

Let required number of machine A and B are x and y respectively.

Since, production of each machine A and B are 60 and 40 units daily respectively.

So, productions by x number of machine A and y number of machine B are 60x and 40y respectively,

Let z denote total output daily, so,

Z = 60x + 40y

Since, each machine of type A and B require 1000 sq.m and 1200 sq.m area so, x machine of type A and y machine of type B require 100x and 1200y sq.m area but, Total area available for machine is 7600 sq.m. so,

$1000\text{x}+1200\text{y}\leq7600$

$5\text{x}+6\text{y}\leq38$ (first constraint)

Since, each machine of type A and B require 12 men and 8 men to work respectively so, x machine of type A and y machine of type B require 12x and By men to work respectively but, Total 72 men available for work so,

$12\text{x}+8\text{y}\leq72$

$3\text{x}+2\text{y}\leq18$ (second constraint)

Hence, mathematical formulation of LPP is,

Find x and y which maximize

Z = 60x + 40y

Subject to constraints,

$5\text{x}+6\text{y}\leq38$

$3\text{x}+2\text{y}\leq18$

$\text{x},\text{y}\geq0$ [Number of machines can not be less than zero]

Region $5\text{x}+6\text{y}\leq38$:

Line 5x + 6y = 38 meets axes at $\text{A}_2\Big(\frac{38}{5},0\Big),\text{B}_1\Big(0,\frac{19}{3}\Big)$ respectively.

Region containing origin represents $5\text{x}+6\text{y}\leq38$ as origin satisfies $5\text{x}+6\text{y}\leq38$.

Region $3\text{x}+2\text{y}\leq18$:

line 3x + 2y = 18 meets axes at A₂(6, 0), B₂(0, 9) respectively.

Region containing origin represents 3x + 2y = 18 as (0, 0) satisfies $3\text{x}+2\text{y}\leq18$.

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

It represents first quadrant.

Shaded region OA₂PB₁, is the feasible region P(4, 3) is obtained by solving 3x + 2y = 18 and 5x+6y - 38

The value of Z = 60x + 40y at

$\text{O}(0, 0) = 60(0) + 40(0) = 0$

$\text{A}_2(6, 0) = 60(6) + 40(0) = 360$

$\text{P}(4,3) =60(4)+ 40 (3) = 360$

$\text{B}_1\Big(0,\frac{19}{3}\Big)=60(0)+ 40\Big(\frac{19}{3}\Big)=\frac{760}{3}$

Therefore maximum 2 = 360 at x = 4, y = 3 or x = 6, y = 0

Output is maximum when 4 machines of type A and 3 machine of type B or 6 machines of type A and no machine of type B.