A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below
|
Type of toys |
Machine I |
Machine II |
Machine III |
|
A |
12 |
18 |
6 |
|
B |
0 |
9 |
6 |
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.
Answer & explanation
Correct answer: option 3
Let x and y toys of type A and B respectively be manufactured in a day.
The given problem can be formulated as follows:
Maximise $z=7.5x+5y$......(1)
subject to the constraints
$2x+y≤60$........(2)
$x≤20$........(3)
$2x+3y≤120$.........(4)
$x,y≥0$..........(5)
The feasible region determined by the constraints is as shown
The corner points of the feasible region are A(20,0),B(20,20),C(15,30) and D(0,40)
The value of z at these corner points are as follows.
|
Corner point |
Z=7.5x+5y |
|
|
A(20,0) |
150 |
|
|
B(20,20) |
150 |
|
|
C(15,30) |
262.5 |
→ Maximum |
|
O(0,40) |
200 |
The maximum value of z is 262.5 at (15,30)
Thus, the manufacturer should manufacture 15 toys of type A and 30 toys of type B to maximize the profit.
