NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

Question:2 Solve the following Linear Programming Problems graphically: Minimise Subject to . Show that the minimum of Z occurs at more than two points

Answer:

The region determined by constraints, is as follows,

The corner points of feasible region are

The value of these points at these corner points are :

Corner points
	6
	-12	Minimum
	0
	16

The minimum value of Z is -12 at

Question:3 Solve the following Linear Programming Problems graphically: Maximise Subject to , , Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, , , is as follows :

The corner points of feasible region are

The value of these points at these corner points are :

Corner points
	9
	0
	10
		Maximum

The maximum value of Z is at

Question:4 Solve the following Linear Programming Problems graphically: Minimise Such that Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints is as follows,

The feasible region is unbounded as shown.

The corner points of the feasible region are

The value of these points at these corner points are :

Corner points
	9
	7	Minimum
	10

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

For this, we draw and check whether resulting half plane has a point in common with the feasible region or not.

We can see a feasible region has no common point with.

Hence, Z has a minimum value of 7 at

Question:5 Solve the following Linear Programming Problems graphically: Maximise Subject to Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, is as follows,

The corner points of feasible region are

The value of these points at these corner points are :

Corner points
	15
	18	Maximum
	10

The maximum value of Z is 18 at

Question:6 Solve the following Linear Programming Problems graphically: Minimise Subject to

Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints is as follows,

The corner points of the feasible region are

The value of these points at these corner points are :

Corner points
	6
	6

Value of Z is the same at both points.

If we take any other point like on line , then Z=6.

Thus the minimum value of Z occurs at more than 2 points .

Therefore, the value of Z is minimum at every point on the line .

Question:7 Solve the following Linear Programming Problems graphically: Minimise and Maximise Subject to Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, is as follows,

The corner points of feasible region are

The value of these points at these corner points are :

Corner points
	400
	600	Maximum
	300	Minimum
	600	maximum

The minimum value of Z is 300 at and maximum value is 600 at all points joing line segment and

Question:8 Solve the following Linear Programming Problems graphically: Minimise and Maximise Subject to Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints is as follows,

The corner points of the feasible region are

The value of these points at these corner points are :

Corner points
	100	Minimum
	100	Minimum
	250
	400	Maximum

The minimum value of Z is 100 at all points on the line segment joining points and .

The maximum value of Z is 400 at .

Question:9 Solve the following Linear Programming Problems graphically: Maximise Subject to the constraints: Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints is as follows,

The corner points of the feasible region are

The value of these points at these corner points are :

Corner points
	- 6	minimum
	-2
	1	maximum

The feasible region is unbounded, therefore 1 may or may not be the maximum value of Z.

For this, we draw and check whether resulting half plane has a point in common with a feasible region or not.

We can see the resulting feasible region has a common point with a feasible region.

Hence , Z =1 is not maximum value , Z has no maximum value.

Question:10 Solve the following Linear Programming Problems graphically: Maximise Subject to Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints is as follows,

There is no feasible region and thus, Z has no maximum value.

NCERT linear programming class 12 solutions - Exercise: 12.2

Question:1 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 Rs 60/kg and Food Q costs Rs.80/kg. Food P contains 3 units/kg of Vitamin A and 5 units / kg of Vitamin B while food Q contains 4 units/kg of Vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.

Answer:

Let mixture contain x kg of food P and y kg of food Q. Thus, .

The given information can be represented in the table as :

The mixture must contain 8 units of Vitamin A and 11 units of Vitamin B.

Therefore, we have

Total cost is Z.

Subject to constraint,

The feasible region determined by constraints is as follows:

It can be seen that a feasible region is unbounded.

The corner points of the feasible region are

The value of Z at corner points is as shown :

corner points
	160	MINIMUM
	160	minimum
	440

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

For this, we draw and check whether resulting half plane has a point in common with the feasible region or not.

We can see a feasible region has no common point with.

Hence, Z has a minimum value 160 at line segment joining points and .

Question:2 One kind of cake requires 200g of flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.

Answer:

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

The given information can be represented in the table as :

Therefore,

.

The total number of cakes, Z. Z=X+Y

Subject to constraint,

The feasible region determined by constraints is as follows:

The corner points of the feasible region are

The value of Z at corner points is as shown :

corner points	Z=X+Y
	25
	30	maximum
	20 0	minimum

The maximum cake can be made 30 (20 of the first kind and 10 of the second kind).

Question:3 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 hour 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 craftsman’s time.

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

Answer:

Let number of rackets be x and number of bats be y.

the machine time availability is not more than 42 hours.

i.e.

craftsman’s time availability is 24 hours

i.e.

The factory has to work at full capacity.

Hence,

Solving equation 1 and 2, we have

Thus, 4 rackets and 12 bats are to be made .

Question:3 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 hour 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 craftsman’s time.

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

Answer:

Let the number of rackets is x and the number of bats is y.

the machine time availability is not more than 42 hours.

craftsman’s time availability is 24 hours

The given information can be repreented in table as shown :

The profit on the bat is 10 and on the racket is 20.

The mathematical formulation is :

maximise

subject to constraints,

The feasible region determined by constraints is as follows:

The corner points are

The value of Z at corner points is as shown :

CORNER POINTS
	160
	200	maximum
	140
	0

Thus, the maximum profit of the factory when it works at full capacity is 200.

Question:4 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 Rs17.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day?

Answer:

Let packages of nuts be x and packages of bolts be y .Thus, .

The given information can be represented in table as :

Profit on a package of nuts is Rs. 17.5 and on package of bolt is 7.

Therefore, constraint are

The feasible region determined by constraints is as follows:

The corner points of feasible region are

The value of Z at corner points is as shown :

Corner points
	70
	73.5	maximum
	28
	0

The maximum value of z is 73.5 at .

Thus, 3 packages of nuts and 3 packages of bolts should be manufactured everyday to get maximum profit.

Question:5 A factory manufactures 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 screws A, while it takes 6 minutes on automatic and 3 minutes on the 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 Rs 7 and screws B at a profit of Rs 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.

Answer:

Let factory manufactures screws of type A and factory manufactures screws of type B. Thus, .

The given information can be represented in the table as :

	screw A	screw B	availability
Automatic machine	4	6
hand operated machine	6	3

Profit on a package of screw A is Rs.7 and on the package of screw B is 10.

Therefore, the constraint is

The feasible region determined by constraints is as follows:

The corner points of the feasible region are

The value of Z at corner points is as shown :

Corner points
	280
	410	maximum
	400
	0

The maximum value of z is 410 at .

Thus, 30 packages of screw A and 20 packages of screw B should be manufactured every day to get maximum profit.

Question:6 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 Rs 5 and that from a shade is Rs 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 maximise his profit?

Answer:

Let the cottage industry manufactures x pedestal lamps and y wooden shades. Thus, .

The given information can be represented in the table as :

Profit on a lamp is Rs. 5 and on the shade is 3.

Therefore, constraint is

The feasible region determined by constraints is as follows:

The corner points of the feasible region are

The value of Z at corner points is as shown :

Corner points
	30
	32	maximum
	30
	0

The maximum value of z is 32 at .

Thus, 4 shades and 4 pedestals lamps should be manufactured every day to get the maximum profit.

Question:7 A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 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 Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit?

Answer:

Let x be Souvenirs of type A and y be Souvenirs of type B .Thus, .

The given information can be represented in table as :

	Type A	Type B	availability
cutting	5	8
asembling	10	8

Profit on type A Souvenirs is Rs. 5 and on type B Souvenirs is 6.

Therefore, constraint are

The feasible region determined by constraints is as follows:

The corner points of feasible region are

The value of Z at corner points is as shown :

Corner points
	120
	160	maximum
	150
	0

The maximum value of z is 160 at .

Thus,8 Souvenirs of type A and 20 Souvenirs of type B should be manufactured everyday to get maximum profit.

Question:8 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 Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000.

Answer:

Let merchant plans has personal computers x desktop model and y portable model

.Thus, .

The cost of desktop model is cost Rs 25000 and portable model is Rs 40000.

Merchant can invest Rs 70 lakhs maximum.

the total monthly demand of computers will not exceed 250 units.

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

Total profit = Z ,

The mathematical formulation of given problem is :

The feasible region determined by constraints is as follows:

The corner points of feasible region are

The value of Z at corner points is as shown :

Corner points
	1125000
	1150000	maximum
	875000
	0

The maximum value of z is 1150000 at .

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

Question:9 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 Rs 4 per unit food and F2 costs Rs 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.

Answer:

Let diet contain x unit of food F1 and y unit of foof F2 .Thus, .

The given information can be represented in table as :

Cost of food F1 is Rs 4 per unit and Cost of food F2 is Rs 6 per unit

Therefore, constraint are

The feasible region determined by constraints is as follows:

We can see feasible region is unbounded.

The corner points of feasible region are

The value of Z at corner points is as shown :

Corner points
	106.67
	104	minimum
	200	maximum

Feasible region is unbounded , therefore 104 may or may not be minimum value of Z .

For this we draw or and check whether resulting half plane has point in common with feasible region or not.

We can see feasible region has no common point with .

Hence , Z has minimum value 104.

Question:10 There are two types of fertilisers 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 atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs 6/kg and F2 costs Rs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Answer:

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

The given information can be represented in table as :

F1 contain 10% nitrogen and F2 contain 5% nitrogen .Farmer requires atleast 14 kg of nitrogen

F1 contain 6% phophoric acid and F2 contain 10% phosphoric acid .Farmer requires atleast 14 kg of nitrogen

Total cost is Z .

Subject to constraint ,

The feasible region determined by constraints is as follows:

It can be seen that feasible region is unbounded.

The corner points of feasible region are

The value of Z at corner points is as shown :

corner points
	1400
	1000	minimum
	1400

Feasible region is unbounded , therefore 1000 may or may not be minimum value of Z .

For this we draw and check whether resulting half plane has point in common with feasible region or not.

We can see feasible region has no common point with .

Hence , Z has minimum value 1000 at point

Question:11 The corner points of the feasible region determined by the following system of linear inequalities:

are and . Let where Condition on p and q so that the maximum of Z occurs at both and is

Answer:

The maximum value of Z is unique.

It is given that maximum value of Z occurs at two points .

Value of Z at =value of Z at

Hence, D is correct option.

NCERT solutions for class 12 maths chapter 12 linear programming-Miscellaneous Exercise

Question:1 Reference of Example 9 (Diet problem): A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol.

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?

Answer:

Let diet contain x packets of food P and y packets of food Q. Thus, .

The mathematical formulation of the given problem is as follows:

Total cost is Z .

Subject to constraint,

The feasible region determined by constraints is as follows:

The corner points of feasible region are

The value of Z at corner points is as shown :

corner points
	150	MINIMUM
	285	maximum
	228

Hence, Z has a maximum value of 285 at the point .

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 is 285 units.

Question:2 A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?

Answer:

Let farmer mix x bags of brand P and y bags of brand Q. Thus, .

The given information can be represented in the table as :

The given problem can be formulated as follows:

Therefore, we have

Subject to constraint,

The feasible region determined by constraints is as follows:

The corner points of the feasible region are

The value of Z at corner points is as shown :

corner points
	4500
	2650
	1950	minimum
	2400

Feasible region is unbounded, therefore 1950 may or may not be a minimum value of Z. For this, we draw and check whether resulting half plane has a point in common with the feasible region or not.

We can see a feasible region has no common point with .

Hence, Z has a minimum value 1950 at point .

Question:3 A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

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

Answer:

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

Mathematical formulation of given problem is as follows:

Minimize :

Subject to constraint ,

The feasible region determined by constraints is as follows:

The corner points of feasible region are

The value of Z at corner points is as shown :

corner points
	160
	112	minimum
	116
	160

The feasible region is unbounded , therefore 112 may or may not be minimum value of Z .

For this we draw and check whether resulting half plane has point in common with feasible region or not.

We can see feasible region has no common point with .

Hence , Z has minimum value 112 at point

Question:4 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:

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

Answer:

Let x and y toys of type A and type B.

Mathematical formulation of given problem is as follows:

Minimize :

Subject to constraint ,

The feasible region determined by constraints is as follows:

The corner points of feasible region are

The value of Z at corner points is as shown :

corner points
	150
	250
	262.5	maximum
	200

Therefore 262.5 may or may not be maximum value of Z .

Hence , Z has maximum value 262.5 at point

Question:5 An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?

Answer:

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

Mathematical formulation of given problem is as follows:

Minimize :

Subject to constraint ,

The feasible region determined by constraints is as follows:

The corner points of feasible region are

The value of Z at corner points is as shown :

corner points
	68000
	136000	maximum
	128000

therefore 136000 is maximum value of Z .

Hence , Z has maximum value 136000 at point

Question:6 Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:

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

Answer:

Let godown A supply x and y quintals of grain to shops D and E respectively. Then , (100-x-y) will be supplied to shop F. Requirements at shop D is 60 since godown A supply x .Therefore remaining (60-x) quintals of grain will be transported from godown B.

Similarly, (50-y) quintals and 40-(100-x-y)=(x+y-60) will be transported from godown B to shop E and F respectively. The problem can be represented diagrammatically as follows:

and

and

Total transportation cost z is given by ,

Mathematical formulation of given problem is as follows:

Minimize :

Subject to constraint ,

The feasible region determined by constraints is as follows:

The corner points of feasible region are

The value of Z at corner points is as shown :

corner points
	560
	620
	610
	510	minimum

therefore 510 may or may not be minimum value of Z .

Hence , Z has miniimum value 510 at point

Question:7 An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is 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 is given in the following table:

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

Answer:

Let x and y litres of oil be supplied from A to petrol pump,D and E. Then , (7000-x-y) will be supplied from A to petrol pump F.

Requirements at petrol pump D is 4500 L. since x L A are transported from depot A,remaining 4500-x L will be transported from petrol pump B

Similarly, (3000-y)L and 3500-(7000-x-y)=(x+y-3500) L will be transported from depot B to petrol E and F respectively.

The problem can be represented diagrammatically as follows:

and

and

Cost of transporting 10 L petrol =Re 1

Cost of transporting 1 L petrol

Total transportation cost z is given by ,

Mathematical formulation of given problem is as follows:

Minimize :

Subject to constraint ,

The feasible region determined by constraints is as follows:

The corner points of feasible region are

The value of Z at corner points is as shown :

corner points
	5000
	5300
	5550
	4400	minimum
	5450

Hence , Z has miniimum value 4400 at point

Question:8 A fruit grower can use two types of fertilizer 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, at least 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 in the garden?

Answer:

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

Mathematical formulation of given problem is as follows:

Minimize :

Subject to constraint ,

The feasible region determined by constraints is as follows:

The corner points of feasible region are

The value of Z at corner points is as shown :

corner points
	595
	550
	470	minimum

Therefore 470 is minimum value of Z .

Hence , Z has minimum value 470 at point

Question:9 Reference of Que 8 : A fruit grower can use two types of fertilizer 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, at least 270 kg of potash and at most 310 kg of chlorine.

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?

Answer:

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

Mathematical formulation of given problem is as follows:

Maximize :

Subject to constraint ,

The feasible region determined by constraints is as follows:

The corner points of feasible region are

The value of Z at corner points is as shown :

corner points
	595	maximum
	550
	470	minimum

therefore 595 is maximum value of Z .

Hence , Z has minimum value 595 at point

Question:10 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 type by at most 600 units. If the company makes profit of Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?

Answer:

Let x and y be number of dolls of type A abd B respectively that are produced per week.

Mathematical formulation of given problem is as follows:

Maximize :

Subject to constraint ,

The feasible region determined by constraints is as follows:

The corner points of feasible region are

The value of Z at corner points is as shown :

corner points
	7200
	15000
	16000	Maximum

Therefore 16000 is maximum value of Z .

Hence , Z has minimum value 16000 at point