Question

A firm manufactures two types of products A and B and sells them at a profit of Rs 5 per unit of type A and Rs 3 per unit of type B. Each product is processed on two machines M₁ and M₂. One unit of type A requires one minute of processing time on M₁ and two minutes of processing time on M₂, whereas one unit of type B requires one minute of processing time on M₁ and one minute on M₂. Machines M₁ and M₂ are respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product should the firm produce a day in order to maximize the profit. Solve the problem graphically.

Answer 1

Let x units of product A and y units of product B were manufactured.
Number of products cannot be negative.
Therefore, $x,y\ge 0$

According to question, the given information can be tabulated as

	Time on ${\mathrm{M}}_1$(minutes)	Time on ${\mathrm{M}}_2$(minutes)
Product A(x)	1	2
Product B(y)	1	1
Availability	300	360

The constraints are
$x+y\le 300$, $2x+y\le 360$

Firm manufactures two types of products A and B and sells them at a profit of Rs 5 per unit of type A and Rs 3 per unit of type B.Therefore, x units of product A and y units of product B costs Rs 5x and Rs 3y respectively.
Total profit = Z = $5x+3y$ which is to be maximised

Thus, the mathematical formulat​ion of the given linear programmimg problem is

Max Z = $5x+3y$

subject to

$x+y\le 300$,
$2x+y\le 360$
$x,y\ge 0$

First we will convert inequations into equations as follows:
x + y = 300, 2x + y = 360, x = 0 and y = 0

Region represented by x + y ≤ 300:
The line x + y = 300 meets the coordinate axes at A₁(300, 0) and B₁(0, 300) respectively. By joining these points we obtain the line
x + y = 30. Clearly (0,0) satisfies the x + y = 30. So, the region which contains the origin represents the solution set of the inequation
x + y ≤ 300.

Region represented by 2x + y ≤ 360:
The line 2x + y = 360 meets the coordinate axes at C₁(180, 0) and D₁(0, 360) respectively. By joining these points we obtain the line
2x + y = 360. Clearly (0,0) satisfies the inequation 2x + y ≤ 360. So,the region which contains the origin represents the solution set of the inequation 2x + y ≤ 360.

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 x + y ≤ 300, 2x + y ≤ 360, x ≥ 0 and y ≥ 0 are as follows.



The corner points are O(0, 0), B₁(0, 300), E₁(60, 240) and C₁(180, 0).

The values of Z at these corner points are as follows

Corner point	Z= 5x + 3y
O	0
B1	900
E1	1020
C1	900

The maximum value of Z is Rs 1020 which is attained at B₁$\left(60,240\right)$.

Thus, the maximum profit is Rs 1020 obtained when 60 units of product A and 240 units of product B were manufactured.