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.  

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Question

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.  

Solution

Correct option is

3 packages & profit Rs 10.50

 

The given information can be summarized in the following tabular form:    

Machines

Time required to produce products

Max. machine hours available

Nut

Bolt

A

B

1

3

3

1

12

12

Profit (in Rs)

2.50

1.00

 

Let the manufacturer produce x packages of nuts and y packages of bolts each day. Since machine A takes one hour to produce one package of nuts and 3 hours to produce one package of bolts. Therefore, the total time required by machine A to produce x packages of nuts and y packages of bolts is (x + 3y) hours. But machine A operates for at most 12 hours.    

∴          x + 3y ≤ 12     

Similarly, the total time required by machine B to produce x packages of nuts and y packages of bolts is (3x + y) hours. But machine B operates for at most 12 hours.   

∴          3x + y ≤ 12 

Since the profit on one package of nuts is Rs 2.50 and on one package of bolts the profit is Rs. 1. Therefore, profit on x packages of nuts and ypackages of bolts is of Rs (2.50x + y). Let Z denote the total profit. Then,Z = 2.50x + y.   

Clearly,    x ≥ 0 and y ≥ 0   

Thus, the above LPP can be stated mathematically as follows:

       Maximize Z = 2.50x + y

Subject to    x + 3y ≤ 12   

                   3x + y ≤ 12   

and,            xy ≥ 0   

To solve this LPP graphically, we first convert the inequations into equations to obtain the following equations.

            x + 3y = 12, 3x + y = 12, x = 0, y = 0   

The line x + 3y = 12 meets the coordinate axes at A1 (12, 0) and B1 (0, 4). Join these two points to obtain the line represented by x + 3y = 12. The region represented by the inequation x + 3y ≤ 12 is the region containing the origin as x = 0, y = 0 satisfies the inequation x + 3y ≤ 12.    

The line 3x + y = 12 meets the coordinate axes at A2 (4, 0) and B2 (0, 12). Join these points to obtain the line represented by 3x + y = 12. Since x = 0,y = 0 satisfies the inequation 3x + y ≤ 12. So, the region containing the origin and below the line 3x + y = 12 represents the region represented by 3x + y ≤ 12.    

Clearly, x ≥ 0 and y ≥ 0 represent all points in first quadrant.   

Thus, the shaded region OA2 PB1 in fig. represents the feasible region of the given LPP.   

The coordinates of the corner-points of the feasible region OA2PB1 are O(0, 0). A2(4, 0), P(3, 3) and B1 (0, 4). These points are obtained by solving the corresponding intersecting lines simultaneously.

                                                                              

 

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

Point (xy)

Value of the objective function

          Z = 2.50x + y

O (0, 0)

A2 (4, 0) 

P (3, 3) 

B1 (0, 4)

Z = 2.50 × 0 + 1 × 0 = 0

Z = 2.50 × 4 + 1 × 0 = 10   

Z = 2.50 × 3 +1 × 3 = 10.50   

Z = 2.50 × 0 + 1 × 4 = 4      

Clearly, Z is maximum at x = 3, y = 3 and the maximum value of Z is 10.50.

Hence, the optimal production strategy  for the manufacturer will be to manufacture 3 packages each of nuts and bolts daily and in this case his maximum profit will be Rs 10.50  

SIMILAR QUESTIONS

Q1

 

Solve the following LPP by graphical method: 

Minimize    Z = 20x + 10y  

Subject to   x + 2y ≤ 40

                   3x + y ≥ 30  

                   4x + 3y ≥ 60 

And,           xy ≥ 0

Q2

 

Solve the following LPP graphically:

Minimize and Maximize Z = 5x + 2y  

Subject to –2x – 3y ≤ – 6  

                     x – 2y ≤ 2

                    3x + 2y ≤ 12  

                  –3x + 2y ≤ 3 

                     xy ≥ 0

Q3

 

Solve the following LPP graphically:

Maximize and Minimize   Z = 3x + 5y  

Subject to   3x – 4y + 12 ≥ 0

                       2x – y + 2 ≥ 0 

                   2x + 3y – 12 ≥ 0 

                               0 ≤ x ≤ 4 

                                      y ≥ 2.

Q4

 

Solve the following linear programming problem graphically:

Maximize  Z = 50x + 15y  

Subject to

            5x + y ≤ 100

             x + y ≤ 60

             xy ≥ 0.

Q5

 

Solve the following LPP graphically:  

Maximize   Z = 5x + 7y  

Subject to

              x + y ≤ 4

             3x + 8y ≤ 24  

            10x + 7y ≤ 35

            xy ≥ 0  

Q6

Solve the following LPP graphically:  

Minimize Z = 3x + 5y     

Subject to  

         – 2x + y ≤ 4  

            x + y ≥ 3

           x – 2y ≤ 2   

           xy ≥ 0

Q7

 

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:

1

2

3

Food Y:

2

2

1

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.

Q8

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. 

Q9

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.    

Q10

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.