A manufacturer has employed 5 skilled men and 10 semi-skilled men and makes two models A and B of an article. The making of one item of model A requires 2 hours of work by a skilled man and 2 hours work by a semi-skilled man. One item of model B requires 1 hour by a skilled man and 3 hours by a semi-skilled man. No man is expected to work more than 8 hours per day. The manufacturer’s profit on an item of model A is Rs. 15 and on an item of model B is Rs 10. How many items of each model should be made per day in order to maximize daily profit? Formulate the above LPP and solve it graphically and find the maximum profit.

 

 

 

 
 
 
 
 

Answers (1)

Let x be the number of items of model A and y be the number of items of model B.

Let z be the required profit

Subject to constraints

    2x + y \leq 8\times 5

\Rightarrow 2x + y \leq 40

2x + 2y \leq 8\times 10 \\ & \Rightarrow 2x + 2y \leq 80

    x \geq 0, \ y\geq 0

Maximize z = 15x + 10y

Changing the above inequalities to equations

2x + y = 40

x 0 20
y 40 0

2x + 3y =80

x 0 40
y 80/3 0

 

Now, plotting these points on the graph

In graph: the shaded region is the required feasible region.

Vertices maximum z = 15x +10y
O(0,0) 15 \times 0 + 10\times 0 = 0
A(20,0) 15 \times 20 + 10\times 0 = 300
B(10,20) 15 \times 10 + 10\times 20 = 350
C(0, \frac{80}{3}) 15 \times 0 + 10\times \frac{80}{3} = 266.6

Maximum profit is obtained when the manufacturer produces 10 items of model A & 20 items of model B and the maximum profit is: 350

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