Question

A furniture trader deals in tables and chairs. He has Rs. 75,000 to invest and a space to store at most 60 items. A table costs him Rs. 1,500 and a chair costs him Rs. 1,000. The trader earns a profit of Rs. 400 and Rs. 250 on a table and chair, respectively. Assuming that he can sell all the items that he can buy, which of the following is/are true for the above problem:

(A) Let the trader buy \( x \) tables and \( y \) chairs. Let \( Z \) denote the total profit. Thus, the mathematical formulation of the given problem is:

\[ Z = 400x + 250y, \]

subject to constraints:

\[ x + y \leq 60, \quad 3x + 2y \leq 150, \quad x \geq 0, \quad y \geq 0. \]

(B) The corner points of the feasible region are (0, 0), (50, 0), (30, 30), and (0, 60).

(C) Maximum profit is Rs. 19,500 when trader purchases 60 chairs only.

(D) Maximum profit is Rs. 20,000 when trader purchases 50 tables only.

Choose the correct answer from the options given below:

• A. (A), (B), and (C) only
• B. (A), (B), and (D) only
• C. (B) and (C) only
• D. (B), (C), and (D) only

Answer 1

The Correct Option is B

To solve the given problem, we need to analyze the mathematical formulation and verify each statement step-by-step:

(A) The problem is formulated correctly.

- Profit function: \( Z = 400x + 250y \)

- Constraints:

• Space constraint: \( x + y \leq 60 \)
• Budget constraint: \( 3x + 2y \leq 150 \)
• Non-negativity constraint: \( x \geq 0, \, y \geq 0 \)

This statement is TRUE.

(B) To find the corner points of the feasible region, solve the following:

- Intersecting lines:

• \( x + y = 60 \) and \( 3x + 2y = 150 \)

Solving simultaneously:

From \( x + y = 60 \), solve for \( y = 60 - x \).

Substitute into the second equation:

\( 3x + 2(60 - x) = 150 \)

\( 3x + 120 - 2x = 150 \, \Rightarrow \, x = 30 \)

Thus, \( y = 60 - 30 = 30 \).

Thus, the corner points are \((0, 0), (50, 0), (30, 30), (0, 60)\).

This statement is TRUE.

(C) Calculate the profit for purchasing 60 chairs only (\( x = 0, y = 60 \)).

Profit: \( Z = 400(0) + 250(60) = 15000 \)

This statement is FALSE. The stated profit is incorrect.

(D) Calculate the profit for purchasing 50 tables only (\( x = 50, y = 0 \)).

Profit: \( Z = 400(50) + 250(0) = 20000 \)

This statement is TRUE. Max profit is indeed Rs. 20,000.

Thus, the correct option is: (A), (B), and (D) only.

Answer 2

The linear programming problem is formulated as:

Z = 400x + 250y,

subject to:

x + y \(≤\) 60, 
3x + 2y \(≤\) 150, 
x ≥ 0, y \(≥\) 0.

The corner points of the feasible region are determined by solving the constraints. They are:

(0, 0), (50, 0), (30, 30), (0, 60).

Calculate Z at each corner point:

Z(50, 0) = 400(50) + 250(0) = 20,000, 
Z(30, 30) = 400(30) + 250(30) = 12,000 + 7,500 = 19,500, 
Z(0, 60) = 400(0) + 250(60) = 15,000.

The maximum profit is Rs. 20,000 when the trader buys 50 tables only.

Thus, statements (A), (B), and (D) are correct.