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.

Model X: 25, Model Y: 30; Max Profit: Rs 40,000

The correct answer is Option (1) → Model X: 25, Model Y: 30; Max Profit: Rs 40,000

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)

$2x + y ≤ 80$   ...(ii)

$x≥0, y ≥0$   ...(iii)

On solving eq. (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.

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