Linear Programming: Overview, Questions, Preparation

EX-12.1

Solve the following Linear Programming Problems graphically:

Q1. Maximize Z = 3x + 4y subject to the constraints: x + y ≤ 4, x ≥ 0, y ≥ 0.

A.1. 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).

 

Q2. Minimise Z = – 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

A.2. 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).

 

Q3. Maximise Z = 5x + 3y subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0

A.3. 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  $\frac{235}{19}at,B\left(\frac{20}{19},\frac{45}{19}\right)$

 

Q4. Minimise Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.

A.4. 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  $B\left(\frac{3}{2},\frac{1}{2}\right)$

 

Q5.  Maximise Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.

A.5. 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

The shaded bounded region OABC in the feasible region with the corner points

$O(0,0),A(5,0),B(4,3),C(0,5)$

The value of Z at these points are given below.

Therefore , the maximum value of Z is 18 at point (4,3).

 

Q6. Minimise Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.

A.6. 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  $x+2y=6$ .

 

Q7. Minimise and Maximise Z = 5x + 10 y subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0.

A.7. 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).

 

Q8.  Minimise and Maximise Z = x + 2y subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; x, y ≥ 0.

A.8. 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).

 

Q9. Maximise Z = – x + 2y, subject to the constraints x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.

A.9. Maximise  $z=-x+2y$ , subject to the constraints

$x\ge 3,x+y\ge 5,x+2y\ge 6,y\ge 0.$

The corresponding equation of the given inequalities are

$x=3$  $x=3$

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

$x+2y=6$  $\frac{x}{6}+\frac{y}{3}=1$

$y=0$  $y=0$

The graph of the given inequalities is shown

The feasible region unbounded.

The values of Z at corner points A(6,0),B(4,1), and C(3,2) are as follows

As the feasible region is unbounded z=1 may or may not be the maximum values.

So, we plot a graph of  $-x+2y>1$

The resulting region has points in common with the feasible region.

Therefore z=1 is not the maximum value. Z has no maximum value.

 

Q10.   Maximise Z = x + y, subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0.

A.10. 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.

 

EXERCISE 12.2

Q1. Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs ₹. 60/kg and Food Q costs ₹. 80/kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while Food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.

A.1. 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.

	Vitamin A (units/kg)	Vitamin B (units/kg)	Cost (Rs/kg)
Food P	3	5	60
Food Q	4	2	80
Requirement (units/kg)	8	11

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

$\begin{array}{ll}3x + 4y \ge 8\hfill & \mathrm{<br>}\hfill \end{array}$

$\begin{array}{l}5x + 2y \ge 11\hfill \end{array}$

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).

 

Q2. One kind of cake requires 200g of flour and 25 g of fat and another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cake which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.

A.2. 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.

	Flour (g)	Fat (g)
Cakes of first kind, x	200	25
Cakes of second kind, y	100	50
Availability	5000	1000

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

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).

 

Q3. 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.

(i) What number of rackets and bats must be made if the factory is to work at full capacity?

(ii) If the profit on a racket and on a bat is Rs. 20 and Rs. 10 respectively, find maximum profit of the factory when it works at full capacity.

A.3. (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.

	Tennis Racket	Cricket Bat	Availability
Machine Time (h)	1.5	3	42
Craftsman’s Time (h)	3	1	24

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

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

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

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,

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

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

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

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.

 

Q4. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of ₹. 17.50 per package on nuts and ₹. 7.00 per package on bolts. How many packages of each should be produced each day so as to maximize his profit, if he operates his machines for at the most 12 hours a day?

A.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.

	Nuts	Bolts	Availability
Machine A (h)	1	3	12
Machine B (h)	3	1	12

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.

 

Q5. A factory manufacturers two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screw A, while it takes 6 minutes on automatic and 3 minutes on hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of ₹. 7 and screws B at a profit of ₹. 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce a day in order to maximize his profit? Determine the maximum profit.

A.5. 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.

	Screw A	Screw B	Availability
Automatic Machine (min)	4	6	4 × 60 =240
Hand Operated Machine (min)	6	3	4 × 60 =240

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.

 

Q6. A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is ₹.5 and that from a shade is ₹.3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximize his profit?

A.6. 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.

	Lamps	Shades	Availability
Grinding/Cutting Machine (h)	2	1	12
Sprayer (h)	3	2	20

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.

 

Q7. A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A requires 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is ₹.5 each for type A and ₹.6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?

A.7. 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.

	Type A	Type B	Availability
Cutting (min)	5	8	3 × 60 + 20 =200
Assembling (min)	10	8	4 × 60 = 240

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.

 

Q8. A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs. 25000 and Rs. 40000 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 ₹ 70 lakhs and if his profit on the desktop model is ₹. 4500 and on portable model is ₹. 5000.

A.8. 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.

Corner point	Z = 4500x + 5000y
A(250, 0)	1125000
B(200, 50)	1150000	→ Maximum
C(0, 175)	875000

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.

 

Q9. A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs ₹. 4 per unit food and F2 costs ₹. 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.

A.9. 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.

	Vitamin A (units)	Mineral (units)	Cost per unit (Rs)
Food F1 (x)	3	4	4
Food F2 (y)	6	3	6
Requirement	80	100

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.

 

Q10. There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs ₹. 6/kg and F2 costs ₹. 5/kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

A.10. 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.

	Nitrogen (%)	Phosphoric Acid (%)	Cost (Rs/kg)
F1 (x)	10	6	6

F2 (y)	5	10	5
Requirement (kg)	14	14

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.

 

Q11. The corner points of the feasible region determined by the following system of linear inequalities:

$2x+y\le 10,x+3y\le 15,x,y\ge 0$ are (0, 0), (5, 0), (3, 4) and (0, 5). Let  $Z=px+qy$ , where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is

(A) p = q

(B) p = 2q

(C) p = 3q

(D) q = 3p

A.11. The maximum value of Z is unique.

It is given that the maximum value of Z occurs at two points, (3, 4) and (0, 5).

∴ Value of Z at (3, 4) = Value of Z at (0, 5)

$\begin{array}{l}\Rightarrow p\left(3\right) + q\left(4\right) = p\left(0\right) + q\left(5\right)\hfill \end{array}$

$\begin{array}{l}\Rightarrow 3p + 4q = 5q\hfill \end{array}$

$\begin{array}{l}\Rightarrow q = 3p\hfill \end{array}$

Hence, the correct answer is D.

Hence option (D) is correct.

 

MISCELLANEOUS EXERCISE

Q1. How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?

A.1. 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:

Corner Point	z = 6x + 3y
A (15, 20)	150
B (40, 15)	285	Maximum
C (2, 72)	228

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.

 

Q2. A farmer mixes two brands, P and Q of cattle feed. Brand P, costing ₹. 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 ₹. 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 for 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?

A.2. 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:

	Vitamin A (units/kg)	Vitamin B (units/kg)	Vitamin C (units/kg)	Cost (Rs/kg)
Food P	3	2.5	2	250
Food Q	1.5	11.25	3	200
Requirement (units/kg)	18	45	24

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:

Corner Point	z = 250x + 200y
A (18, 0)	4500
B (9, 2)	2650
C (3, 6)	1950	Minimum
D (0, 12)	2400

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.

 

Q3. A dietician wishes to mix together two kinds of food, X and Y, in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg of food are given below:

Food	Vitamin A	Vitamin B	Vitamin C
X	1	2	3
Y	2	2	1

One kg of food X costs ₹. 16, and one kg of food Y costs ₹. 20. Find the least cost of the mixture which will produce the required diet.

A.3. 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,

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

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

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

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

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:

Corner Point	z = 16x + 20y
A (10, 0)	160
B (2, 4)	112	Minimum

C (1, 5)	116
D (0, 8)	160

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.

 

Q4. A manufacturer makes two types of toys, A and B. Three machines are needed for this purpose, and the time (in minutes) required for each toy on the machines is given below:

Types of Toys	Machines
	I	II	III
A	12	18	6
B	6	0	9

Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is ₹. 7.50 and that on each toy of type B is ₹. 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.

A.4. Let x and y toys of type A and type B be manufactured in a day, respectively.

The given problem can be formulated as given below:

$Maximisez=7.5x+5y\dots \dots \dots \dots ..\left(i\right)$

Subject to the constraints,

$\begin{array}{}\end{array}2x+y\le 60\dots \dots \dots \dots .\left(ii\right)$

$x\le 20\dots \dots \dots ..\left(iii\right)$

$2x+3y\le 120\dots \dots \dots ..\left(iv\right)$

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

The feasible region determined by the constraints is given below:

A (20, 0), B (20, 20), C (15, 30) and D (0, 40) are the corner points of the feasible region.

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

Corner Point	z = 7.5x + 5y
A (20, 0)	150
B (20, 20)	250
C (15, 30)	262.5	Maximum
D (0, 40)	200

262.5 at (15, 30) is the maximum value of z.

Hence, the manufacturer should manufacture 15 toys of type A and 30 toys of type B to maximise the profit.

 

Q5. An aeroplane can carry a maximum of 200 passengers. A profit of ₹. 1000 is made on each executive class ticket, and a profit of ₹. 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?

A.5. 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:

Corner Point	z = 1000x + 600y
A (20, 80)	68000
B (40, 160)	136000	Maximum
C (20, 180)	128000

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.

 

Q6. Two godowns, A and B, have grain capacities 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 is given in the following table:

Transportation Cost per Quintal (in Rs)
From/To	A	B
D	6	4
E	3	2
F	2.50	3

How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?

A.6. 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:

Corner Point	z = 2.5x + 1.5y + 410
A (60, 0)	560
B (60, 40)	620
C (50, 50)	610
D (10, 50)	510	Minimum

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.

 

Q7. An oil company has two depots, A and B, with capacities of 7000 L and 4000 L, respectively. The company has to supply oil to three petrol pumps, D, E and F, whose requirements are 4500L, 3000L and 3500L, respectively. The distances (in km) between the depots and the petrol pumps are given in the following table:

Distance in (km)
From/To	A	B
D	7	3
E	6	4
F	3	2

Assuming that the transportation cost of 10 litres of oil is ₹. 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?

A.7. 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:

Corner Point	z = 0.3x + 0.1y + 3950
A (3500, 0)	5000
B (4500, 0)	5300
C (4500, 2500)	5550
D (4000, 3000)	5450
E (500, 3000)	4400	Minimum

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.

 

Q8. A fruit grower can use two types of fertilisers in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table, Tests indicate that the garden needs at least 240 kg of phosphoric acid, 270 kg of potash and at most 310 kg of chlorine.

If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added to the garden?

Kg per Bag
	Brand P	Brand Q
Nitrogen	3	3.5
Phosphoric acid	1	2
Potash	3	1.5
Chlorine	1.5	2

A.8. 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:

Corner Point	z = 3x + 3.5y
A (140, 50)	595
B (20, 140)	550
C (40, 100)	470	Minimum

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.

 

Q9. Refer to Question 8. If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?

A.9. 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:

Corner Point	z = 3x + 3.5y
A (140, 50)	595	Maximum
B (20, 140)	550
C (40, 100)	470

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.

 

Q10. A toy company manufactures two types of dolls, A and B. Market research and available resources have indicated that the combined production level should not exceed 1200 dolls per week, and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other types by at most 600 units. If the company makes a profit of ₹. 12 and ₹. 16 per doll on dolls A and B, respectively, how many of each should be produced weekly in order to maximise the profit?

A.10. 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,

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

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

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

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

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:

Corner Point	z = 12x + 16y
A (600, 0)	7200
B (1050, 150)	15000
C (800, 400)	16000	Maximum

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.