A company manufactures two types of product A and B. Each unit of A requires 3 grams of nickle and 1 gram of chromium, while each unit of B requires 1 gram of nickle and 2 grams of chromium. The firm can produce 9 grams of nickle and 8 grams of chromium. The profit is ₹40 on each unit of product A and ₹50 on each unit of type B. How many unit of each type should the company manufacture so as to earn maximum profit? Use linear programming to find the solution.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2 units of product A and 3 units of product B, for a profit of ₹230.
Let x and y be the number of products of types A and B respectively manufactured by the company. As the profit on each product of type A is ₹40 and on each product of type B is ₹50, so the total profit is $Z = 40x + 50y$ (in ₹).
Hence, the problem can be formulated as an L.P.P. as follows:
Maximize $Z = 40x + 50y$ subject to the constraints
$3x + 1y ≤9$ (nickle constraint)
$1x + 2y ≤8$ (chromium constraint)
$x ≥0, y ≥0$ (number of product cannot be negative)
We draw the straight lines $3x + y = 9$ and $x + 2y = 8$ and shade the region statisfied by the inequalities.
The shaded portion shows the feasible region which is bounded.
The point of intersection of the lines $3x + y = 9$ and $x + 2y = 8$ is $B (2, 3)$.
The corner points of the feasible region OABC are O(0, 0), A(3, 0), B(2, 3) and C(0, 4). The optimal solution occurs at one of the corner points.
At $O(0, 0), Z = 40 × 0 + 50 × 0 = 0$.
At $A(3, 0), Z = 40 × 3 + 50 × 0 = 120$.
At $B(2, 3), Z = 40 × 2 + 50 × 3 = 230$.
At $C(0, 4), Z = 40 × 0 + 50 × 4 = 200$.
We find that the value of Z is maximum at B (2, 3).
Hence, the company should manufacture 2 products of type A and 3 products of type B to get a maximum profit of ₹230.