Question

A company produces two types of goods, A and B, that require gold and silver. Each unit of type A requires 3 gm of silver and 1 gm of gold while that of type B requires 1 gm of silver and 2 gm of gold. The company can produce 9 gm of silver and 8 gm of gold. If each unit of type A brings a profit of Rs 40 and that of type B Rs 50, find the number of units of each type that the company should produce to maximize the profit. What is the maximum profit?

Answer 1

Let x goods of type A and y goods of type B were produced.
Number of goods cannot be negative.
Therefore, $x,y\ge 0$

The given information can be tabulated as follows:

	Silver(gm)	Gold white(gm)
Type A	3	1
Type B	1	2
Availability	9	8

Therefore, the constraints are
$$\begin{gathered} 3x+y\le 9 \\ x+2y\le 8 \end{gathered}$$

If each unit of type A brings a profit of Rs 40 and that of type B Rs 50.Then, x goods of type A and y goods of type B brings a profit of Rs 40x and Rs 50y.
Total profit = Z = $40x+50y$ which is to be maximised.

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

Max Z = $40x+50y$

subject to

$$\begin{gathered} 3x+y\le 9 \\ x+2y\le 8 \end{gathered}$$
$x,y\ge 0$

First we will convert inequations into equations as follows :
3x + y = 9, x + 2y = 8, x = 0 and y = 0

Region represented by 3x + y ≤ 9:
The line 3x + y = 9 meets the coordinate axes at A₁(3, 0) and B₁(0, 9) respectively. By joining these points we obtain the line
3x + y = 9.Clearly (0,0) satisfies the 3x + y = 9. So,the region which contains the origin represents the solution set of the inequation
3x + y ≤ 9.

Region represented by x + 2y ≤ 8:
The line x + 2y = 8 meets the coordinate axes at C₁(8, 0) and D₁(0, 4) respectively. By joining these points we obtain the line
x + 2y = 8.Clearly (0,0) satisfies the inequation x + 2y ≤ 8. So,the region which contains the origin represents the solution set of the inequation x + 2y ≤ 8.

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 3x + y ≤ 9, x + 2y ≤ 8, x ≥ 0, and y ≥ 0 are as follows.



The corner points are O(0, 0), D₁(0, 4), E₁(2, 3), A₁(3, 0)

The values of Z at these corner points are as follows

Corner point	Z = 40x + 50y
O	0
D1	200
E1	230
A1	120

The maximum value of Z is 230 which is attained at E₁(2, 3).

Thus, the maximum profit is of Rs 230 obtained when 2 units of type A and 3 units of type B produced.