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Linear Programming — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsLinear Programming5 Marks
Question
A manufacturer produces two products A and B. Both the products are processed on two different machines. The available capacity of first machine is 12 hours and that of second machine is 9 hours per day. Each unit of product A requires 3 hours on both machines and each unit of product B requires 2 hours on first machine and 1 hour on second machine. Each unit of product A is sold at Rs. 7 profit and that of B at a profit of Rs. 4. Find the production level per day for maximum profit graphically.
✓

Answer

Let x units of product A and y units of product B be manufactured by the manufacturer per day.
It is given that one unit of product A requires 3 hours of processing time on first machine, while one unit of product B requires 2 hours of processing time on first machine.
It is also given that first machine is available for 12 hours per day.
$\therefore$ 3x + 2y ≤ 12
Also, one unit of product A requires 3 hours of processing time on second machine, while one unit of product B requires 1 hour of processing time on second machine.
It is also given that second machine is available for 9 hours per day.
$\therefore$ 3x + y ≤ 9
The profits on one unit each of product A and product B is Rs. 7 and Rs 4, respectively.
So, the objective function is given by Z = Rs. (7x + 4y).
Therefore, the mathematical formulation of the given linear programming problem can be stated as:

Maximize Z = 7x + 4y
Subject to the constraints
3x + 2y ≤ 12 .....(1)
3x + y ≤ 9 .....(2)
x ≥ 0, y ≥ 0 .....(3)
The feasible region determined by constraints (1) to (3) is graphically represented as:

Here, it is seen that OABCO is the feasible region and it is bounded.
The values of Z at the corner points of the feasible region are represented in tabular form as:
Corner point
Z = 7x + 4y
O(0, 0)
Z = 7 × 0 + 4 × 0 = 0
A(0, 10)
Z = 7 × 3 + 4 × 0 = 21
B(173, 0)
Z = 7 × 2 + 4 × 3 = 26
C(3, 8)
Z = 7 × 0 + 4 × 6 = 24
The maximum value of Z is 26, which is obtained at x = 2 and y = 3.
Thus, 2 units of product A and 3 units of product B should be manufactured by the manufacturer per day in order to maximize the profit.
Also, the maximum daily profit of the manufacturer is Rs. 26.

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