Question

An airplane can carry a maximum of \(250\) passengers. A profit of Rs \(1500\) is made on each executive class ticket and a profit of Rs \(900\) is made on each economy class ticket. The airline reserves at least \(30\) seats for executive class. However at least \(4\) times as many passengers prefer to travel by economy class than by executive class. Let \(x_1\) be the number of passengers of executive class and \(x_2\) be the number of passengers of economy class. Formulate the LPP in order to maximize the profit for the airline...

• A. Maximize \(z = 1500x_1+900x_2\) subject to \(x_1+x_2\leq 250\), \(x_1\leq 30\), \(x_2\leq 4x_1\), \(x_1\geq 0,x_2\geq 0\).
• B. Minimize \(z = 150x_1+90x_2\) subject to \(x_1+x_2\leq 250\), \(x_1\geq 30\), \(x_2\geq 4x_1\), \(x_1\geq 0,x_2\geq 0\).
• C. Minimize \(z = 1500x_1+900x_2\) subject to \(x_1+x_2\leq 250\), \(x_1\geq 30\), \(x_2\geq 4x_1\), \(x_1\geq 0,x_2\geq 0\).
• D. Maximize \(z = 1500x_1+900x_2\) subject to \(x_1+x_2\leq 250\), \(x_1\geq 30\), \(x_2\geq 4x_1\), \(x_1\geq 0,x_2\geq 0\).

Hint:

Maximise profit, with at least 30 executive seats, at least four times as many economy as executive passengers, and at most 250 passengers.

Answer 1

The Correct Option is D

Step 1: Define the variables
Let \(x_1\) be the number of executive class passengers and \(x_2\) the number of economy class passengers.

Step 2: Objective
Profit is Rs 1500 per executive ticket and Rs 900 per economy ticket. The airline wants the largest profit, so we maximise \(z = 1500x_1 + 900x_2\).

Step 3: Constraints
Capacity: \(x_1 + x_2 \le 250\). At least 30 executive seats: \(x_1 \ge 30\). At least 4 times as many prefer economy as executive: \(x_2 \ge 4x_1\). Non-negativity: \(x_1, x_2 \ge 0\).

Step 4: Match the options
This is option (D). Option (A) has the inequalities reversed for \(x_1\) and \(x_2\). Options (B) and (C) say "Minimize", which does not fit a profit problem.

Final Answer:
Option (D) states the correct formulation. This is option (D). \[ \boxed{\text{(D) }\text{Maximize } z=1500x_1+900x_2} \]