Question

A manufacturer produces two Models of bikes-Model X and Model Y. Model X takes a 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hour available per week. Handling and Marketing costs are Rs 2000 and Rs 1000 per unit for Models X and Y respectively. The total funds available for these purposes are Rs 80,000 per week. Profits per unit for Models X and Y are Rs 1000 and Rs 500, respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.

Answer 1

Let x and y be the number of Models of bike produced by the manufacturer.

Given information is

Model X takes 6 man-hours to make per unit

Model Y takes 10 man-hours to make per unit

Total man-hours available = 450

∴ 6x + 10y ≤ 450

⇒ 3x + 5y ≤ 225  ......(i)

Handling and marketing cost of Model X and Y are ₹ 2,000 and ₹ 1,000 respectively

Total funds available is ₹ 80,000 per week

∴ 2000x + 1000y ≤ 80,000

⇒ 2x + y ≤ 80   ......(ii)

And x ≥ 80, y ≥ 0

Profit (Z) per unit of models X and Y are ₹ 1,000 and ₹ 500 respectively

So, Z = 1000x + 500y

The required LPP is

Maximise Z = 1000x + 500y subject to the constraints

3x + 5y ≤ 225  .......(i)

x	0	75
y	45	0

2x + y ≤ 80  ......(ii)

x	0	40
y	80	0

x ≥ 0, y ≥ 0  ......(iii)

On solving equation (i) and (ii)

We get, x = 25, y = 30

Here, the feasible region is OABC

Whose corner points are O(0, 0), A(40, 0), B(25, 30) and C(0, 45).

Let us evaluate the value of Z.

Corner points	Value of Z = 1000x + 500y
O(0, 0)	Z = 0 + 0 = 0
A(40, 0)	Z = 1000(40) + 0 = 40,000	← Maximum
B(25, 30)	Z = 1000(25) + 500(30) = 40,000	← Maximum
C(0, 45)	Z = 0 + 500(45) = 22500

Hence, the maximum profit is ₹ 40,000 by producing 25 bikes of Model X and 30 bikes of Model Y.