Question

A company sells two different products, A and B. The two products are produced in a common production process, which has a total capacity of 500 man-hours. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The market has been surveyed and company officials feel that the maximum number of unit of A that can be sold is 70 and that for B is 125. If the profit is Rs 20 per unit for the product A and Rs 15 per unit for the product B, how many units of each product should be sold to maximize profit?

Answer 1

Let x units of product A and y units of product B were manufactured.
Clearly, $x\ge 0,y\ge 0$
It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B.The two products are produced in a common production process, which has a total capacity of 500 man-hours.

$5x+3y\le 500$

The maximum number of unit of A that can be sold is 70 and that for B is 125.

$$\begin{gathered} x\le 70 \\ y\le 125 \end{gathered}$$

If the profit is Rs 20 per unit for the product A and Rs 15 per unit for the product B. Therefore, profit x units of product A and y units of product B is Rs 20x and Rs 15y respectively.

Total profit = Z = $20x+15y$

The mathematical formulation of the given problem is

Max Z = $20x+15y$
subject to

$$\begin{gathered} 5x+3y\le 500 \\ x\le 70 \\ y\le 125 \end{gathered}$$
$x\ge 0,y\ge 0$

First we will convert inequations into equations as follows:
5x + 3y = 500, x = 70, y = 125, x = 0 and y = 0

Region represented by 5x + 3y ≤ 500:
The line 5x + 3y = 500 meets the coordinate axes at A₁(100, 0) and $B_1\left(0,\frac{500}{3}\right)$ respectively. By joining these points we obtain the line 5x + 3y = 500. Clearly (0,0) satisfies the 5x + 3y = 500. So, the region which contains the origin represents the solution set of the inequation 5x + 3y ≤ 500.

Region represented by x ​ ≤ 70:
The line ​x = 70 is the line passes through C₁(70, 0) and is parallel to Y axis. The region to the left of the line x = 70 will satisfy the inequation x ​ ≤ 70.

Region represented by y ​ ≤ 125:
The line ​y = 125 is the line passes through D₁(0, 125) and is parallel to X axis. The region below the the line y = 125 will satisfy the inequation y ​ ≤ 125.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.

The feasible region determined by the system of constraints 5x + 3y ≤ 500, x ​ ≤ 70, y ​ ≤ 125, x ≥ 0 and y ≥ 0 are as follows.


The corner points are O(0, 0), D₁$\left(0,125\right)$, E₁(25, 125), F₁(70, 50) and C₁(70, 0).

The values of Z at the corner points are

Corner points	Z = $20x+15y$
O	0
D1	1875
E1	2375
F1	2150
C1	1400

The maximum value of Z is 2375 which is at E₁$\left(25,125\right)$.

Thus, maximum profit is Rs 2375, 25 units of A and 125 units of B should be manufactured.