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Linear Programming

Question
CBSEENMA12033565

A manufacturer has three machines 1, II and III installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must be operated for at least 5 hours a day. She produces only two items M and N each requiring the use of all the three machines.
The number of hours required for producing 1 unit of each of M and N on the three machines are given in the following table:

Items

Number of 1

ours required

on machines

 

I

II

III

M

1

2

1

N

2

1

1.25

She makes a profit of Rs. 600 and Rs. 400 on items M and N respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit?

Solution

Let x and y be the number of items M and N respectively.
Total profit on the production = Rs. (600 x + 400 y)
Let Z be the profit.
Mathematical formulation of the given problem is as follows:
Maximise Z = 600 x + 400 y
subject to the constraints
                straight x space plus space 2 straight y space less or equal than space 12
2 straight x space plus space straight y space less or equal than space 12
straight x plus 5 over 4 straight y space greater or equal than space 5
straight x greater or equal than 0 comma space straight y greater or equal than 0

Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + 2y = 12
For x = 0, 2 y = 12 or y = 6
For y = 0, x = 12
∴ line meets OX in A(12, 0) and OY in L(0, 6).
Also we draw the graph of 2x + y = 12
For x = 0, y = 12
For y = 0, 2x = 12 or x = 6
∴ line meets OX in B(6, 0) and OY in M(0, 12).
Again we draw the graph of 
                          straight x plus 5 over 4 straight y space equals space 5
For   x = 0,  5 over 4 straight y space equals space 5 space space or space space straight y space equals space 4
For  y = 0,  x = 5
∴ line meets OX in C(5, 0) and OY in N(0, 4).
Since feasible region satisfies all the constraints.
∴ CBDLN is the feasible region.

The corner points are C(5, 0), B(6, 0),
D(4, 4), L(0, 6), N(0, 4).
At C(5, 0), Z = 600 × 5 + 400 × 0 = 3000 + 0 = 3000
At B(6, 0), Z = 600 × 6 + 400 × 0 = 3600 + 0 = 3600
At D(4, 4), Z = 600 × 4 + 400 × 4 = 2400 + 1600 = 4000
At L(0, 6), Z = 600 × 0 + 400 × 6 = 0 + 2400 = 2400
At N(0, 4), Z = 600 × Ot 400 × 4 = 0 + 1600 = 1600
∴ maximum value = 4000 at (4, 4).
∴ manufacturer has to produce 4 units of each item to get the maximum profit of Rs. 4000.