Question

Refer to question 11. How many of circuits of Type A and of Type B, should be produced by the manufacturer so as to maximise his profit? Determine the maximum profit.

Answer 1

As per the solution of Question No. 11

We have Maximise Z = 50x + 60y

Subject to the contraints

2x + y ≤ 20  ......(i)

x + 2y ≤ 12  .......(ii)

x + 3y ≤ 15   .......(iv)

x ≥ 0, y ≥ 0  .......(iv)

Let us draw the table for the above statements

Table for (i)

x	0	10
y	20	0

Table for (ii)

x	0	10
y	20	0

Table for (ii)

x	0	15
y	5	0

Solving equation (i) and (ii) we get,

x = `28/3`

y = `4/3`

∴ `"B"(38/3, 4/3)` is the corner

Solving equation (ii) and (iii) we get,

x = 6, y = 3

∴ C(6, 3) is the corner

Solving equation (i) and (iii) we get,

x = 9, y = 2  .....(Not included in the feasible region)

Here, OABCD is the feasible region.

So, the corner points are O(0, 0), A(10, 0), `"B"(28/3, 4/3)`, C(6, 3) and D(0, 5).

Let us evaluate the value of Z

Corner points	Corresponding values of Z = 50x + 60y
O(0, 0)	Z = 50(0) + 60(0) = 0
A(10, 0)	Z = 50(10) + 60(0) = 500
`"B"(28/3, 4/3)`	Z = `50(28/3) + 60(4/3)` = `1400/3 + 240/3`  = `1640/3` = 546.6	← Maximum
C(6, 3)	Z = 50(6) + 60(3) = 480
D(0, 5)	Z = 50(0) + 60(5) = 300

Here, the maximum profit is ₹ 546.6 which is not possible for number of items in fraction.

Hence, the maximum profit for the manufacturer is ₹ 480 at (6, 3).

Type A = 6 and Type B = 3.