A dealer wishes to purchase a number of fans and sewing machines. He has only Rs 5760.00 to invest and has space for at most 20 items. A fan costs him Rs 360.00 and a sewing machine Rs 240.00. His expectation is that he can sell a fan at a profit of Rs 22.00 and a sewing machine at a profit of Rs 18.00. Assuming that he can sell all the items that he can buy, how should he invest his money in order to maximize his profit? Translate this problem mathematically and then solve it. 


Correct option is

x = 8 and y = 12 & maximum value of Z is 392.


Suppose the dealer buys x fans and y sewing machines. Since the dealer has space for at most 20 items. Therefore,   

           x + y ≤ 20   

A fan costs Rs 360 and a sewing machine costs Rs 240. Therefore, total cost of x fans and y sewing machines is Rs (360x + 240y). But the dealer has only Rs 5760 to invest. Therefore, 

          360x + 240y ≤ 5760 

Since the dealer can sell all the items that he can buy and the profit on a fan is of Rs 22 and on a sewing machine the profit is of Rs 18. Therefore, total profit on selling x fans and y sewing machines is of Rs (22x + 18y).  

Let Z denote the total profit. Then, Z = 22x + 18y.  

Clearly, xy ≥ 0.  

Thus, the mathematical formulation of the given problem is

          Maximize Z = 22x + 18y

Subject to   

            x + y ≤ 20

      360x + 240y ≤ 5760 

and, x ≥ 0, y ≥ 0  

To solve this LPP graphically, we first covert the inequations into equations and draw the corresponding lines. The feasible region of the LPP is shaded in fig. The corner points of the feasible region OA2 PB1 areO (0, 0), A2 (16, 0), P (8, 12) and B1 (0, 20).



These points have been obtained by solving the corresponding intersecting lines, simultaneously.  

The values of the objective function Z at corner-points of the feasible region are given in the following table.

Point (xy)

Value of the objective function

              Z = 22x + 18y

O (0, 0)

Z = 22 × 0 + 18 × 0 = 0

A2 (16, 0)

Z = 22 × 16 + 18 × 0 = 352

(8, 12)

Z = 22 × 8 + 18 × 12 = 392

B1 (0, 20)

Z = 22 × 0 + 20 × 18 = 360

Clearly, Z is maximum at x = 8 and y = 12. The maximum value of Z is 392.

Hence, the dealer should purchase 8 fans and 12 sewing machines to obtain the maximum profit under given conditions.




Solve the following linear programming problem graphically:

Maximize  Z = 50x + 15y  

Subject to

            5x + y ≤ 100

             x + y ≤ 60

             xy ≥ 0.



Solve the following LPP graphically:  

Maximize   Z = 5x + 7y  

Subject to

              x + y ≤ 4

             3x + 8y ≤ 24  

            10x + 7y ≤ 35

            xy ≥ 0  


Solve the following LPP graphically:  

Minimize Z = 3x + 5y     

Subject to  

         – 2x + y ≤ 4  

            x + y ≥ 3

           x – 2y ≤ 2   

           xy ≥ 0



A house wife 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 is given below:   


Vitamin A

Vitamin B

Vitamin C

Food X:




Food Y:




One kg of food X costs Rs 6 and one kg of food Y costs Rs 10. Find the least cost of the mixture which will produce the diet.


A dietician wishes to mix two types of food in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food ‘I’ contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C while food ‘II’ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs 5.00 per kg to purchase food ‘I’ and Rs 7.00 per kg to produce food ‘II’. Determine the minimum cost to such a mixture. formulate the above as a LPP and solve it. 


Every gram of wheat provides 0.1 gm of proteins and 0.25 gm of carbohydrates. The corresponding values of rice are 0.05 gm and 0.5 gm respectively. Wheat costs Rs. 4 per kg and rice Rs. 6. The minimum daily requirements of proteins and carbohydrates for an average child are 50 gms and 200 gms respectively. In what quantities should wheat and rice be mixed in the daily diet to provide minimum daily requirements of proteins and carbohydrates at minimum cost.    


A manufacturer produces nuts and bolts for industrial machinery. It takes 1 hour or work on machine A and 3 hours on machine B to produce a package of nuts while it takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs 2.50 per package of nuts and Re 1.00 per package of bolts. How many packages or each should he produce each day so as to maximize hit profit, if he operates his machines for at most 12 hours a day? Formulate this mathematically and then solve it.  


An oil company requires 12,000, 20,000 and 15,000 barrels of high-grade, medium grade and low grade oil, respectively. Refinery A produces 100, 300 and 200 barrels per day of high-grade, medium-grade and low-grade oil, respectively, while refinery B produces 200, 400 and 100 barrels per day of high-grade, medium-grade and low-grade oil, respectively. If refinery A costs Rs 400 per day and refinery B  costs Rs 300 per day to operate, how many days should each be run to minimize costs while satisfying requirements.


A company produces soft drinks that has a contract which requires that a minimum of 80 units of the chemical A and 60 units of the chemical B to go into each bottle of the drink. The chemicals are available in a prepared mix from two different suppliers. Supplier S has a mix of 4 units of A and 2 units of B that costs Rs 10, the supplier T has a mix of 1 unit of A and 1 unit of B that costs Rs 4. How many mixes from S and T should the company purchase to honour contract requirement and yet minimize cost?


A farm is engaged in breeding pigs. The pigs are fed on various products grown on the farm. In view of the need to ensure certain nutrient constituents (call them XY and Z). it is necessary to buy two additional products, say A and B. One unit of product A contains 36 units of X, 3 units of Y, and 20 units of Z. One unit of product B contains 6 units of X, 12 units of Y and 10 units of Z. The minimum requirement of XY and Z is 108 units, 36 units and 100 units respectively. Product A costs Rs 20 per unit and product B costs Rs 40 per unit. Formulate the above as a linear programming problem to minimize the total cost, and solve the problem by using graphical method.