A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) requires for each toy on the machines is given below:
|
Types of Toys |
Machines |
||
|
I |
II |
III |
|
|
A |
12 |
18 |
6 |
|
B |
6 |
0 |
9 |
Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs. 7.50 and that on each toy of type B is Rs. 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
Let P be the profit.
Table
We are to maximise
subject to the constraints
12 x + 6 y ≤ 360 or 2 x + y ≤ 60
18 x ≤ 360 or x ≤ 20
6 x + 9 y ≤ 360 or 2 x + 3 y ≤ 120
x ≥ 0, y ≥ 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 2x + y = 60
For x = 0, y = 60
For y = 0, 2x = 60
or x = 30
∴ line meets OX in A(30, 0) and OY in L(0, 60).
Now x = 20 is a straight line BM as parallel to y-axis at a distance of 20.
Again we draw the graph of 2 x + 3 y = 120
For x = 0, 3 y = 120 or y = 40
For y = 0, 2 x = 120 or x = 60
∴ line meets OX in C(60, 0) and OY in N(0, 40).
Since feasible region satisfies all the constraints.
∴ OBDEN is the feasible region.
The corner points are O(0, 0), B(20, 0), D(20, 20), E(15, 30). N(0, 40).
∴ maximum value = 262.5 at (15, 30).
∴ maximum profit is Rs. 262.5 when 15 toys of type A and 30 toys of type B are manufactured.
