Question

A firm manufactures two products A and B. Each product is processed on two machines M₁ and M₂. Product A requires 4 minutes of processing time on M₁ and 8 min. on M₂ ; product B requires 4 minutes on M₁ and 4 min. on M₂. The machine M₁ is available for not more than 8 hrs 20 min. while machine M₂ is available for 10 hrs. during any working day. The products A and B are sold at a profit of Rs 3 and Rs 4 respectively.
Formulate the problem as a linear programming problem and find how many products of each type should be produced by the firm each day in order to get maximum profit.

Answer 1

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

The given information can be tabulated as follows:

Product	$M_1$(minutes)	$M_2$(minutes)
A(x)	4	8
B(y)	4	4
Availability	500	600

Therefore, the constraints are

$$\begin{gathered} 4x+4y\le 500 \\ 8x+4y\le 600 \end{gathered}$$

The products A and B are sold at a profit of Rs 3 and Rs 4 respectively. Therefore, Profit gained from x products of type A and y products of type B is Rs 3x and Rs 4y respectively.
Total profit = Z = 3x + 4y which is to be maximised.

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

Max Z = $3x+4y$

subject to

$$\begin{gathered} 4x+4y\le 500 \\ 8x+4y\le 600 \end{gathered}$$

First we will convert inequations into equations as follows:
4x + 4y = 500, 8x + 4y = 600, x = 0 and y = 0

Region represented by 4x + 4y ≤ 500:
The line 4x + 4y = 500 meets the coordinate axes at A₁(125, 0) and B₁(0, 125) respectively. By joining these points we obtain the line
4x + 4y = 500. Clearly (0,0) satisfies the 5x + 20y = 400 . So, the region which contains the origin represents the solution set of the inequation 5x + 20y ≤ 400.

Region represented by 8x + 4y ≤ 600:
The line 8x + 4y = 600 meets the coordinate axes at C₁(75, 0) and D₁(0, 150) respectively. By joining these points we obtain the line
8x + 4y = 600. Clearly (0,0) satisfies the inequation 8x + 4y ≤ 600. So,the region which contains the origin represents the solution set of the inequation
8x + 4y ≤ 600.

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 4x + 4y ≤ 500, 8x + 4y ≤ 600, x ≥ 0, and y ≥ 0 are as follows.



The corner points are O(0, 0), B₁(0, 125), E₁(25, 100) and C₁(75, 0).

The values of Z at these corner points are as follows

Corner point	Z= 3x + 4y
O	0
B1	500
E1	475
C1	225

The maximum value of Z is 500 which is attained at B₁(0, 125).
Thus, the maximum profit is Rs 500 obtained when no units of product A and 125 units of product B were manufactured.