Question

A manufacturer produces two types of products - A and B, which are subjected to two types of operations, viz. grinding and polishing.
- Each unit of product A takes 2 hours of grinding and 3 hours of polishing.
- Each unit of product B takes 3 hours of grinding and 2 hours of polishing.
The manufacturer has 10 grinders and 15 polishers. Each grinder operates for 12 hours per day and each polisher for 10 hours per day.
The profit margin per unit of A and B are Rs. 5 and Rs. 7 respectively.
If the manufacturer utilises all his resources for producing these two types of items, what is the maximum profit that the manufacturer can earn in a day?

• A. Rs. 280
• B. Rs. 294
• C. Rs. 515
• D. Rs. 550
• E. None of the above

Hint:

In linear programming type problems, always write the constraints carefully, then solve the simultaneous equations to find feasible solutions. Finally, substitute values into the profit function to maximize it.

Answer 1

The Correct Option is B

Step 1: Define variables.
Let the number of units of product A be \(x\).
Let the number of units of product B be \(y\).
Step 2: Write grinding and polishing constraints.
- Each unit of A requires 2 hours of grinding and 3 hours of polishing.
- Each unit of B requires 3 hours of grinding and 2 hours of polishing.
Hence:
Grinding hours required = \(2x + 3y\).
Polishing hours required = \(3x + 2y\).
Step 3: Calculate available resources.
Number of grinders = 10, each working 12 hours/day = \(10 \times 12 = 120\) hours available per day.
Number of polishers = 15, each working 10 hours/day = \(15 \times 10 = 150\) hours available per day.
So:
Grinding constraint = \(2x + 3y = 120\). \hfill (1)
Polishing constraint = \(3x + 2y = 150\). \hfill (2)
Step 4: Solve the simultaneous equations.
Multiply equation (1) by 3: \(6x + 9y = 360\).
Multiply equation (2) by 2: \(6x + 4y = 300\).
Subtract: \((6x+9y) - (6x+4y) = 360 - 300\).
\(\;\;\Rightarrow 5y = 60\).
\(\;\;\Rightarrow y = 12\).
Substitute into equation (1): \(2x + 3(12) = 120\).
\(\;\;\Rightarrow 2x + 36 = 120\).
\(\;\;\Rightarrow 2x = 84\).
\(\;\;\Rightarrow x = 42\).
So the optimal solution is \(x = 42\) units of A and \(y = 12\) units of B.
Step 5: Calculate profit.
Profit = 5 per unit of A + 7 per unit of B.
= \(42 \times 5 + 12 \times 7\).
= 210 + 84.
= Rs. 294.
Step 6: Conclude.
The maximum profit that the manufacturer can earn in a day is Rs. 294. \[ \boxed{294 \; \text{(Option B)}} \]