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?
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
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
For x = 0,
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.