Question

A carpenter earns a profit of ₹50 and ₹80 on one chair and one table respectively. The requirement and availability of wood and labour are tabled as:

Required	Chair	Table	Available Quantity
Wood Labour	3 1	5 2	150 56

The number of chairs and tables in appropriate units to be manufactured for maximum profit are, respectively:

• A. 0, 28
• B. 50,0
• C. 20, 18
• D. 0,30

Answer 1

The Correct Option is B

To determine the number of chairs and tables that maximize profit under the given constraints, let's define variables and use linear programming:

Let \( x \) be the number of chairs and \( y \) be the number of tables.

Objective Function: Maximize Profit \( P = 50x + 80y \)

Subject to Constraints:

• Wood: \( 3x + 5y \leq 150 \)
• Labour: \( x + 2y \leq 56 \)
• Non-negativity: \( x \geq 0, y \geq 0 \)

The constraints form a system of inequalities. We will solve these systematically for potential solutions.

1. From Labour Constraint: \( x + 2y \leq 56 \)

Solve for \( x \): \( x = 56 - 2y \)

2. From Wood Constraint: \( 3x + 5y \leq 150 \)

Replace \( x \) from Step 1: \( 3(56 - 2y) + 5y \leq 150 \)

Simplify: \( 168 - 6y + 5y \leq 150 \)

Rearrange: \( -y \leq -18 \) → \( y \geq 18 \)

Combining \( x = 56 - 2y \) with \( y \geq 18 \):

When \( y = 18, x = 56 - 36 = 20 \), satisfying the conditions.

Now evaluate possible extreme points of the feasible region for maximum profit:

Testing options, solving equations:

a) \( x = 50, y = 0 \) → Profit \( P = 50(50) + 80(0) = 2500 \)

b) \( x = 0, y = 30 \) → Check if \( y=30 \leq 28 \) from \( y \geq 18 \): Invalid

c) \( x = 20, y = 18 \) → Profit \( P = 50(20) + 80(18) = 2440 \)

The largest profit comes from \( x = 50, y = 0 \).

Thus, the number of chairs and tables to maximize profit is 50, 0.