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?
Solution
Let x goods of type A and y goods of type B were produced.
Number of goods cannot be negative.
Therefore, \[x, y \geq 0\]
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
\[3x + y \leq 9\]
\[x + 2y \leq 8\]
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 formulation of the given linear programmimg problem is
Max Z = \[40x + 50y\]
subject to
\[3x + y \leq 9\]
\[x + 2y \leq 8\]
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.
