Class 12 Maths Chapter 12 Linear Programming NCERT Solutions PDF

Let the cottage industry manufacture x pedestal lamps and y wooden shades. Therefore,

x ≥ 0 and y ≥ 0

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

The profit on a lamp is Rs 5 and on the shades is Rs 3. Therefore, the constraints are

$\begin{array}{l}2x + y \le 12\hfill \end{array}$

$\begin{array}{l}3x + 2y \le 20\hfill \end{array}$

$\begin{array}{l}Total profit, Z = 5x + 3y\hfill \end{array}$

The mathematical formulation of the given problem is

Maximize $Z=5x +3y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}2x + y \le 12 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}3x + 2y \le 20 \dots \left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 0 \dots \left(4\right)\hfill \end{array}$

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

The corner points are A (6, 0), B (4, 4), and C (0, 10).

The values of Z at these corner points are as follows

The maximum value of Z is 32 at (4, 4).

Thus, the manufacturer should produce 4 pedestal lamps and 4 wooden shades to maximize his profits.

Let there be x cakes of first kind and y cakes of second kind. Therefore,

x ≥ 0 and y ≥ 0

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

$\begin{array}{l}\therefore 200x+100y\le 5000\\ \Rightarrow 2x+y\le 50\\ 25x+50y\le 1000\\ \Rightarrow x+2y\le 40\end{array}$

Total numbers of cakes, Z, that can be made are, $Z= x + y$

The mathematical formulation of the given problem is

Maximize $Z= x + y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}2x+y\le 50.......(2)\\ x+2y\le 40.......(3)\\ x,y\ge 0..............(4)\end{array}$

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

The corner points are A (25, 0), B (20, 10), O (0, 0), and C (0, 20).

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

Thus, the maximum numbers of cakes that can be made are 30 (20 of one kind and 10 of the other kind).

Let x and y litres of oil be supplied from A to the petrol pumps, D and E. So, $\left(7000–x–y\right)$ 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 $,\left(3000–y\right)Land3500–\left(7000–x–y\right)=\left(x+y–3500\right)$ L will be transported from depot B to petrol pumps E and F, respectively.

The given problem can be represented diagrammatically as given below:

$\begin{array}{}\end{array}x\ge 0,y\ge 0,and\left(7000–x–y\right)\ge 0$

$Then,x\ge 0,y\ge 0,andx+y\le 7000$

$4500–x\ge 0,3000–y\ge 0,andx+y–3500\ge 0$

$Then,x\le 4500,y\le 3000,andx+y\ge 3500$

Cost of transporting 10 L of petrol = Rs. 1

Cost of transporting 1 L of petrol = Rs. 1/10

Hence, the total transportation cost is given by,

$\begin{array}{l}z = \left(7/10\right) x + \left(6/10\right) y + 3 / 10 \left(7000–x–y\right) + 3 / 10 \left(4500–x\right) + 4 / 10 \left(3000–y\right) + 2 / 10 \left(x+y–3500\right)\hfill \end{array}$

$\begin{array}{l}= 0.3x + 0.1y + 3950\hfill \end{array}$

The problem can be formulated as given below:

$Minimisez=0.3x+0.1y+3950\dots \dots \dots .\left(i\right)$

Subject to constraints,

$\begin{array}{}\end{array}x+y\le 7000\dots \dots \dots .\left(ii\right)$

$x\le 4500\dots \dots \dots ..\left(iii\right)$

$y\le 3000\dots \dots .\left(iv\right)$

$x+y\ge 3500\dots \dots \dots \left(v\right)$

$x,y\ge 0\dots \dots \dots \dots \left(vi\right)$

The feasible region determined by the constraints is given below:

A (3500, 0), B (4500, 0), C (4500, 2500), D (4000, 3000) and E (500, 3000) are the corner points of the feasible region.

The values of z at these corner points are given below:

The minimum value of z is 4400 at (500, 3000).

Hence, 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.

Therefore, the minimum transportation cost is ?. 4400.

Let the farmer mix x bags of brand P and y bags of brand Q, respectively

The given information can be compiled in a table as given below:

The given problem can be formulated as given below:

$\begin{array}{}\end{array}Minimisez=250x+200y\dots \dots \dots \dots \left(i\right)$

$3x+1.5y\ge 18\dots \dots \dots \dots ..\left(ii\right)$

$2.5x+11.25y\ge 45\dots \dots \dots ..\left(iii\right)$

$2x+3y\ge 24\dots \dots \dots \dots ..\left(iv\right)$

$x,y\ge 0\dots \dots \dots .\left(v\right)$

The feasible region determined by the system of constraints is given below:

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

The values of z at these corner points are given below:

Here, the feasible region is unbounded; hence, 1950 may or may not be the minimum value of z.

For this purpose, we draw a graph of the inequality, $250x+200y<1950or5x+4y<39$ , and check whether the resulting half-plane has points in common with the feasible region or not.

Here, it can be seen that the feasible region has no common point with $5x+4y<39$ .

Hence, at point (3, 6), the minimum value of z is 1950.

Therefore, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimise the cost to ?. 1950.

Let the farmer buy x kg of fertilizer F₁ and y kg of fertilizer F₂. Therefore,

x ≥ 0 and y ≥ 0

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

F₁ consists of 10% nitrogen and F₂ consists of 5% nitrogen. However, the farmer requires at least 14 kg of nitrogen.

$\therefore 10\mathrm{\%}of x +5\mathrm{\%}of y \ge 14$

$\begin{array}{l}\frac{x}{10}+\frac{y}{20}\ge 14\\ 2x+y\ge 280\end{array}$

F₁ consists of 6% phosphoric acid and F₂  consists of 10% phosphoric acid. However, the farmer requires at least 14 kg of phosphoric acid.

$\therefore 6\mathrm{\%}of x +10\mathrm{\%}of y \ge 14$

$\begin{array}{l}\frac{6x}{100}+\frac{10y}{100}\ge 14\\ 3x+56y\ge 700\end{array}$

Total cost of fertilizers, $Z=6x +5y$

The mathematical formulation of the given problem is

Minimize $Z=6x +5y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}2x + y \ge 280 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}3x + 5y \ge 700 \dots \left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 0 \dots \left(4\right)\hfill \end{array}$

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

It can be seen that the feasible region is unbounded.

The corner points are  $A\left(\frac{700}{3},0\right),B\left(100,80\right),C\left(0,280\right)$ .

The values of Z at these points are as follows.

As the feasible region is unbounded, therefore, 1000 may or may not be the minimum value of Z.

For this, we draw a graph of the inequality, $6x +5y <1000$ , and check whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with

$6x +5y <1000$

Therefore, 100 kg of fertiliser F1 and 80 kg of fertilizer F2 should be used to minimize the cost. The minimum cost is ? 1000.

Minimize $z=x+2y$

Subject to $2x+y\ge 3,x+2y\ge 6,x,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}2x+y=3\\ x+2y=6\\ x,y\ge 0\end{array}$

$\Rightarrow \frac{x}{\frac{3}{2}}+\frac{y}{3}=1$

$\frac{x}{6}+\frac{y}{3}=1$

$x,y\ge 0$

The feasible region is unbounded the corner point are A (6,0), B (0,3)

The value of Z at these corner points are follows.

Since, the feasible region is unbounded, a graph of $x+2y<6$ is drawn.

Also since there is no point common in feasible region and region $x+2y<6$ .

$z=6$ is maximum on all points joining line (0,3), (6,0)

i.e,  $z=6$ will be minimum on $x+2y=6$ 

Minimize and Maximise $z=5x+10y$

Subject to $x+2y\le 120,x+y\ge 60,x-2y\ge 0,x,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}x+2y=120\\ x+y=60\\ x-2y=0\\ x,y=0\end{array}$

$\Rightarrow \frac{x}{120}+\frac{y}{60}=120$

$\begin{array}{l}\frac{x}{60}+\frac{y}{60}=1\\ x=2y\\ x,y=0\end{array}$

The graph of the inequalities in shown.

The shaded founded region ABCD is the feasible region with the corner points $A\left(60,0\right),B\left(120,0\right),C\left(60,30\right)\&D\left(40,20\right)$

The value of Z a these corner points are

The minimum value of Z is 300 at (60,0) and the maximum value of Z is 600 at all the points on the line segment joining B (120,0) and C (60,30).

Let the fruit grower use x bags of brand P and y bags of brand Q, respectively.

The problem can be formulated as given below:

$Minimisez=3x+3.5y\dots \dots \dots \dots .\left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}x+2y \ge 240\dots \dots \dots ..\left(ii\right)$

$x+0.5y\ge 90\dots \dots ..\left(iii\right)$

$1.5x+2y\le 310\dots \dots \dots ..\left(iv\right)$

$x,y\ge 0\dots \dots \dots \dots .\left(v\right)$

The feasible region determined by the system of constraints is given below:

A (240, 0), B (140, 50) and C (20, 140) are the corner points of the feasible region.

The values of z at these corner points are given below:

The maximum value of z is 470 at (40, 100).

Therefore, 40 bags of brand P and 100 bags of brand Q should be added to the garden to minimise the amount of nitrogen.

Hence, the minimum amount of nitrogen added to the garden is 470 kg.

Let the merchant stock x desktop models and y portable models. Therefore,

x ≥ 0 and y ≥ 0

The cost of a desktop model is Rs 25000 and of a portable model is Rs 4000. However, the merchant can invest a maximum of Rs 70 lakhs.

$\begin{array}{l}\therefore 25000x + 40000y \le 7000000\hfill \end{array}$

$\begin{array}{l}5x +8y \le 1400\hfill \end{array}$

The monthly demand of computers will not exceed 250 units.

$\therefore x+y\le 250$

The profit on a desktop model is Rs 4500 and the profit on a portable model is Rs 5000.

Total profit, $Z=4500x +5000y$

Thus, the mathematical formulation of the given problem is

Maximum $Z=4500x+5000y .....\left(1\right)$

subject to the constraints,

$\begin{array}{l}5x +8y \le 1400 ....\left(2\right)\hfill \end{array}$

$\begin{array}{l}\therefore x + y \le 250 .....\left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 01400 ......\left(4\right)\hfill \end{array}$

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

The corner points are A (250, 0), B (200, 50), and C (0, 175).

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

The maximum value of Z is 1150000 at (200, 50).

Thus, the merchant should stock 200 desktop models and 50 portable models to get the maximum profit of ? 1150000.

Let the diet contain x units of food F₁ and y units of food F₂. Therefore,

x ≥ 0 and y ≥ 0

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

The cost of food F₁ is Rs 4 per unit and of Food F₂  is ? 6 per unit. Therefore, the constraints are

$\begin{array}{}\end{array}3x +6y \ge 80$

$4x +3y \ge 100$

$x, y \ge 0$

$Totalcostofthediet,Z=4x +6y$

The mathematical formulation of the given problem is

Minimise $Z=4x +6y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}3x + 6y \ge 80 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}4x + 3y \ge 100 \dots \left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 0 \dots \left(4\right)\hfill \end{array}$

The feasible region determined by the constraints is as follows.

It can be seen that the feasible region is unbounded.

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

The corner points are $A\left(\frac{80}{3},0\right),B\left(24,\frac{4}{3}\right),C\left(0,\frac{100}{3}\right)$ .

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

As the feasible region is unbounded, therefore, 104 may or may not be the minimum value of Z.

For this, we draw a graph of the inequality, $4x +6y <104or2x +3y <52$ , and check whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with $2x +3y <52$

Therefore, the minimum cost of the mixture will be ? 104.

Let godown A supply x and y quintals of grain to shops D and E.

So, $\left(100–x–y\right)$ will be supplied to shop F.

Since x quintals are transported from godown A, the requirement at shop D is 60 quintals. Hence, the remaining (60 – x) quintals will be transported from godown B.

Similarly, (50 – y) quintals and $40–\left(100–x–y\right)=\left(x+y–60\right)$ quintals will be transported from godown B to shop E and F.

The given problem can be represented diagrammatically as given below:

$x\ge 0,y\ge 0,and100–x–y\ge 0$

Then, $x\ge 0,y\ge 0,andx+y\le 100$

$\begin{array}{l}60 – x \ge 0, 50 – y \ge 0, and x + y – 60 \ge 0\hfill \end{array}$

$\begin{array}{l}Then, x \le 60, y \le 50, and x + y \ge 60\hfill \end{array}$

Total transportation cost z is given by,

$\begin{array}{l}z = 6x + 3y + 2.5 \left(100–x–y\right) + 4 \left(60–x\right) + 2 \left(50–y\right) + 3 \left(x+y–60\right)\hfill \end{array}$

$\begin{array}{l}= 6x + 3y + 250 – 2.5x – 2.5y + 240 – 4x + 100 – 2y + 3x + 3y – 180\hfill \end{array}$

$\begin{array}{l}= 2.5x + 1.5y + 410\hfill \end{array}$

The given problem can be formulated as given below:

$Minimisez=2.5x+1.5y+410\dots \dots \dots \dots .\left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}x+y\le 100\dots \dots \dots ..\left(ii\right)$

$x\le 60\dots \dots \dots ..\left(iii\right)$

$y\le 50\dots \dots \dots .\left(iv\right)$

$x+y\ge 60\dots \dots \dots \left(v\right)$

$x,y\ge 0\dots \dots \dots \dots ..\left(vi\right)$

The feasible region determined by the system of constraints is given below:

A (60, 0), B (60, 40), C (50, 50) and D (10, 50) are the corner points.

The values of z at these corner points are given below:

The minimum value of z is 510 at (10, 50).

Hence, 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 are 50 quintals, 0 quintals, 0 quintals, respectively.

Thus, the minimum cost is Rs. 510.

Maximise $z=3x+4y$

Subject to the constraints: $x+y\le 4,x\ge 0,y\ge 0$

The corresponding equation of the above inequality are

$\begin{array}{l}x+y=4\\ x=0,y=0\end{array}$

$\Rightarrow \frac{x}{4}+\frac{y}{4}=1$

$x=0,y=0$

The graph of the given inequalities.

The shaded region OAB is the feasible region which is bounded.

The corresponding of the corner point of the feasible region are O (0,0), A (4,0), and B (0,4).

The value of Z at these points are as follows,

Corner point $z=3x+4y$

O (0,0) 0

A (4,0) 12

B (0,4) 16

Therefore the maximum value of Z is 16 at the point B (0,4).

Let the manufacturer produce x packages of nuts and y packages of bolts. Therefore,

x ≥ 0 and y ≥ 0

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

The profit on a package of nuts is Rs 17.50 and on a package of bolts is Rs 7. Therefore, the constraints are

$\begin{array}{l}x + 3y \le 12 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}3x + y \le 12 \dots (3\hfill \end{array}$

Total profit, $Z=17.5x +7y$

The mathematical formulation of the given problem is

Maximise $Z=17.5x +7y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}x + 3y \le 12 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}3x + y \le 12 \dots \left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 0 \dots \left(4\right)\hfill \end{array}$

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

The corner points are A (4, 0), B (3, 3), and C (0, 4).

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

The maximum value of Z is ? 73.50 at (3, 3).

Thus, 3 packages of nuts and 3 packages of bolts should be produced each day to get the maximum profit of ? 73.50.

Let the mixture contain x kg of food X and y kg of food Y, respectively.

The mathematical formulation of the given problem can be written as given below:

$Minimisez=16x+20y\dots \dots ..\left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}x+2y\ge 10\dots \dots \dots \dots .\left(ii\right)$

$x+y\ge 6\dots \dots \dots \dots \left(iii\right)$

$3x+y\ge 8\dots \dots \dots \dots .\left(iv\right)$

$x,y\ge 0\dots \dots \dots \dots \left(v\right)$

The feasible region determined by the system of constraints is given below:

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

The values of z at these corner points are given below:

Since the feasible region is unbounded, 112 may or may not be the minimum value of z.

For this purpose, we draw a graph of the inequality, $16x+20y<112or4x+5y<28$ , and check whether the resulting half-plane has points in common with the feasible region or not.

Here, it can be seen that the feasible region has no common point with $4x+5y<28$ .

Hence, the minimum value of z is 112 at (2, 4).

Therefore, the mixture should contain 2 kg of food X and 4 kg of food Y.

The minimum cost of the mixture is ? 112.

Let x and y be the number of dolls of type A and B, respectively, that are produced in a week.

The given problem can be formulated as given below:

$Maximisez=12x+16y\dots \dots \dots ..\left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}x+y \le 1200\dots \dots \dots \dots \left(ii\right)$

$y\le x/2orx\ge 2y\dots \dots \dots .(ii\mathrm{i)}$

$x–3y\le 600\dots \dots \dots \dots .\left(iv\right)$

$x,y\ge 0\dots \dots \dots \dots \dots \left(v\right)$

The feasible region determined by the system of constraints is given below:

A (600, 0), B (1050, 150) and C (800, 400) are the corner points of the feasible region.

The values of z at these corner points are given below:

The maximum value of z is 16000 at (800, 400).

Hence, 800 and 400 dolls of type A and type B should be produced, respectively, to get the maximum profit of? 16000.

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

x ≥ 0 and y ≥ 0

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

$\begin{array}{l}4x + 6y \le 240\hfill \end{array}$

$\begin{array}{l}6x + 3y \le 240\hfill \end{array}$

Total profit, $Z=7x +10y$

The mathematical formulation of the given problem is

Maximize $Z=7x +10y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}4x + 6y \le 240 ....\left(2\right)\hfill \end{array}$

$\begin{array}{l}6x + 3y \le 240 .....\left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 0 \dots \left(4\right)\hfill \end{array}$

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 ? 410.

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

x ≥ 0 and y ≥ 0

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

$\begin{array}{l} 5x + 8y \le 200\hfill \end{array}$

$\begin{array}{l}10x + 8y \le 240 i.e.,5x + 4y \le 120\hfill \end{array}$

$\begin{array}{l}Total profit, Z = 5x + 6y\hfill \end{array}$

The mathematical formulation of the given problem is

Maximize $Z=5x +6y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l} 5x + 8y \le 200 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}5x + 4y \le 120 \dots \left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 0 \dots \left(4\right)\hfill \end{array}$

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

The corner points 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 ? 160.

Let the airline sell x tickets of executive class and y tickets of economy class, respectively.

The mathematical formulation of the given problem can be written as given below:

$Maximisez=1000x+600y\dots \dots \dots \dots \left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}x+y \le 200\dots \dots \dots \dots \dots ..\left(ii\right)$

$x\ge 20\dots \dots \dots \dots \left(iii\right)$

$y–4x\ge 0\dots \dots \dots \dots \dots \left(iv\right)$

$x,y\ge 0\dots \dots \dots \dots \dots \left(v\right)$

The feasible region determined by the constraints is given below:

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

The values of z at these corner points are given below:

136000 at (40, 160) is the maximum value of z.

Therefore, 40 tickets of the executive class and 160 tickets of the economy class should be sold to maximise the profit, and the maximum profit is? 136000.

Let the mixture contain x kg of food P and y kg of food Q. Therefore, x ≥ 0 and y ≥ 0

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

The mixture must contain at least 8 units of vitamin A and 11 units of vitamin B. Therefore, the constraints are

$3x + 4y \ge 8$

$5x + 2y \ge 11$

Total cost, Z, of purchasing food is, $Z=60x +80y$

The mathematical formulation of the given problem is

Minimise $Z=60x +80y \dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}3x + 4y \ge 8 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}5x + 2y \ge 11 \dots \left(3\right)\hfill \end{array}$

$\begin{array}{l}x, y \ge 0 \dots \left(4\right)\hfill \end{array}$

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

It can be seen that the feasible region is unbounded.

The corner points of the feasible region are A(8/3,0) ,B(2,1/2) and C(0,11/2)

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

As the feasible region is unbounded, therefore, 160 may or may not be the minimum value of Z.

For this, we graph the inequality, $60x +80y <160or3x +4y <8$ , and check whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with $3x +4y <8$

Therefore, the minimum cost of the mixture will be ? 160 at the line segment joining the points (8/3,0) and (2,1/2).

(i) Let the number of rackets and the number of bats to be made be x and y respectively.

The machine time is not available for more than 42 hours.

$\therefore 1.5x+3y\le 42....\left(1\right)$

The craftsman’s time is not available for more than 24 hours.

$\therefore 3x+y\le 24 ......\left(2\right)$

The factory is to work at full capacity. Therefore,

$\begin{array}{l}1.5x + 3y = 42\hfill \end{array}$

$\begin{array}{l}3x + y = 24\hfill \end{array}$

On solving these equations, we obtain

x = 4 and y = 12

Thus, 4 rackets and 12 bats must be made.

(i) The given information can be complied in a table as follows.

$\begin{array}{l}\therefore 1.5x + 3y \le 42\hfill \end{array}$

$\begin{array}{l}3x + y \le 24\hfill \end{array}$

$\begin{array}{l}x, y \ge 0\hfill \end{array}$

The profit on a racket is Rs 20 and on a bat is Rs 10.

$\therefore Z=20x+10y$

The mathematical formulation of the given problem is

Maximize $Z=20x+10y\dots \left(1\right)$

subject to the constraints,

$\begin{array}{l}1.5x + 3y \le 42 \dots \left(2\right)\hfill \end{array}$

$\begin{array}{l}3x + y \le 24 \dots (\mathrm{3)}\hfill \end{array}$

$\begin{array}{ll}\hfill & x, y \ge 0 \dots \left(4\right)\hfill \end{array}$

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

The corner points are A (8, 0), B (4, 12), C (0, 14), and O (0, 0).

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

Thus, the maximum profit of the factory when it works to its full capacity is Rs 200.

Let the diet contain x and y packets of foods P and Q, respectively. Hence,

x ≥ 0 and y ≥ 0

The mathematical formulation of the given problem is given below:

$Maximisez=6x+3y\dots \dots \dots \dots ..\left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}4x+y\ge 80\dots \dots \dots \dots .\left(ii\right)$

$x+5y\ge 115\dots \dots \dots .\left(iii\right)$

$3x+2y\le 150\dots \dots \dots \dots \left(iv\right)$

$x,y\ge 0\dots \dots \dots \dots \dots \left(v\right)$

The feasible region determined by the system of constraints is given below:

A (15, 20), B (40, 15) and C (2, 72) are the corner points of the feasible region

The values of z at these corner points are as given below:

So, the maximum value of z is 285 at (40, 15).

Hence, to maximise the amount of vitamin A in the diet, 40 packets of food P and 15 packets of food Q should be used.

The maximum amount of vitamin A in the diet is 285 units.

Maximise $z=x+y$ , subject to $x-y\le -1,-x+y\le 0,x,y\ge 0$

The corresponding equation of the given inequalities are

$x-y=1$ $\frac{x}{-1}+\frac{y}{1}=1$

$-x+y=0$ $\Rightarrow$ $x=y$

$x,y=0$ $x,y=0$

The graph of the given inequalities is shown.

There is no common point in the two shaded region. Thus, there is no feasible region.

$\therefore$ Z has no maximum value.

Minimize and Maximise $z=x+2y$

Subject to $x+2y\ge 100,2x-y\le 0,2x+y\le 200,x,y\ge 0$

The corresponding equation of the given inequalities are

$x+2y=100$ $\Rightarrow \frac{x}{100}+\frac{y}{50}=1$

$2x-y=0$ $2x=y$

$2x+y=200$ $\frac{x}{100}+\frac{y}{200}=1$

$x,y\ge 0$ $x,y=0$

The graph of the inequalities is shown below.

The shaded bounded region ABCD is the feasible region with the corner points.

A (0,50), B, (20,40), C (50,100), D (0,200)

The values of Z at these corner points are

The maximum value of Z is 400 at D (0,200) and the minimum value of Z is 100 at all the points on the line segment joining the points A (0,50) and B (20,40).

Let the fruit grower use x bags of brand P and y bags of brand Q, respectively.

The problem can be formulated as given below:

$Maximisez=3x+3.5y\dots \dots \dots ..\left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}x+2y \ge 240\dots \dots \dots ..\left(ii\right)$

$x+0.5y\ge 90\dots \dots \dots ..\left(iii\right)$

$1.5x+2y\le 310\dots \dots \dots ..\left(iv\right)$

$x,y\ge 0\dots \dots \dots \dots .\left(v\right)$

The feasible region determined by the system of constraints is given below:

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

The values of z at these corner points are given below:

The maximum value of z is 595 at (140, 50).

Hence, 140 bags of brand P and 50 bags of brand Q should be used to maximise the amount of nitrogen.

Thus, the maximum amount of nitrogen added to the garden is 595 kg.

Minimize $z=-3x+4y$

Subject to $x+2y\le 8,3x+2y\le 12,x\ge 0,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}x+2y=8\\ 3x+2y=12\\ x=0,y=0\end{array}$

$\Rightarrow \frac{x}{8}+\frac{y}{4}=1$

$\begin{array}{l}\frac{x}{4}+\frac{y}{6}=1\\ x=0,y=0\end{array}$

The graph is shown below.

The bounded region OABC is the feasible region with the corner points O (0,0), A (4,0), B (2,3), and C (0,4

The value of Z at these points are

Therefore, the minimum value of Z is -12 at (4,0).

Maximise $z=5x+3y$

Subject to $2x+5y\le 15,5x+2y\le 10,x\ge 0,y\ge 0$

The corresponding equation of the above linear inequalities are

$\begin{array}{l}3x+5y=15\\ 5x+2y=10\&x=0,y=0\end{array}$

$\Rightarrow \frac{x}{5}+\frac{y}{3}=1$

$\begin{array}{l}\frac{x}{2}+\frac{y}{5}=1\\ x=0,y=0\end{array}$

The graph of its given inequalities.

The shaded region OABC is the feasible region which is bounded with the corner points

$O\left(0,0\right),A\left(2,0\right),B\left(\frac{20}{19},\frac{45}{19}\right)\&C\left(0,3\right)$

The values of Z at these points are

Therefore the maximum value of Z is $\frac{235}{19}at,B\left(\frac{20}{19},\frac{45}{19}\right)$

Minimize $z=3x+5y$

Such that $x+3y\ge 3,x+y\ge 2,x,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}x+3y=3\\ x+y=2\\ x,y=0\end{array}$

$\begin{array}{l}\Rightarrow \frac{x}{3}+\frac{y}{1}=1\\ \frac{x}{2}+\frac{y}{2}=1\\ x,y=0\end{array}$

The graph of the given inequalities is

The feasible region is unbounded. The corner points are $A\left(3,0\right),B\left(\frac{3}{2},\frac{1}{2}\right)\&C\left(0,2\right)$

The values of Z at these corner points as follows.

As the feasible region is unbounded, 7 may or may not be minimum value of Z.

We draw the graph of inequality $3x+5y<7$ .

The feasible region has no common point with $3x+5y<7$ .

Therefore minimum value of Z in 7 at $B\left(\frac{3}{2},\frac{1}{2}\right)$

Maximum $z=3x+2y$

Subject to $x+2y\le 10,3x+y=\le 15,x,y\ge 0$

The corresponding equation of the given inequalities are :

$\begin{array}{l}x+2y=10\\ 3x+y=15\\ x,y=0\end{array}$

$\Rightarrow \frac{x}{10}+\frac{y}{5}=1$

$\begin{array}{l}\frac{x}{5}+\frac{y}{5}=1\\ x,y=0\end{array}$

The graph of the given inequalities