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 address the problem, we will analyze the mathematical formulation and verify each statement methodically:

(A) The problem formulation is accurate.

- Objective Function: \( Z = 400x + 250y \)

- Constraints:

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

This statement is TRUE.

(B) The corner points of the feasible region are determined by solving the following:

- Intersection of constraint lines:

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

Solving the system:

From \( x + y = 60 \), we get \( y = 60 - x \).

Substituting into the budget constraint:

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

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

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

The corner points identified are \((0, 0), (50, 0), (30, 30), (0, 60)\).

This statement is TRUE.

(C) Profit calculation for purchasing 60 chairs exclusively (\( x = 0, y = 60 \)).

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

This statement is FALSE. The stated profit is erroneous.

(D) Profit calculation for purchasing 50 tables exclusively (\( x = 50, y = 0 \)).

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

This statement is TRUE. The maximum profit is indeed Rs. 20,000.

Therefore, the correct options are: (A), (B), and (D) only.