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: Read the sentence by sentence
"Maximum of 250" gives an upper limit on the total. "At least 30" gives a lower limit on $x_1$. "At least 4 times as many" gives $x_2 \ge 4x_1$.

Step 2: Direction of optimisation
Profit is to be maximised, so the objective is $z = 1500x_1 + 900x_2$ with the Rs 1500 and Rs 900 as stated.

Step 3: Check feasibility
The point $x_1 = 30$, $x_2 = 120$ satisfies all constraints ($150 \le 250$, $120 \ge 120$). So the feasible region is not empty.

Step 4: Reject
Option (B) changes the profit to 150 and 90 and minimises. Option (C) minimises. Option (A) wrongly caps $x_1$ at 30 and $x_2$ at $4x_1$.

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