Q80 / 150 · MHT CET 2026 (19 April Shift 2) · Mathematics
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...
- AMaximize $z = 1500x_1 + 900x_2$ subject to $x_1 + x_2 \leq 250$, $x_1 \leq 30$, $x_2 \leq 4x_1$, $x_1 \geq 0, x_2 \geq 0$.
- BMinimize $z = 150x_1 + 90x_2$ subject to $x_1 + x_2 \leq 250$, $x_1 \geq 30$, $x_2 \geq 4x_1$, $x_1 \geq 0, x_2 \geq 0$.
- CMinimize $z = 1500x_1 + 900x_2$ subject to $x_1 + x_2 \leq 250$, $x_1 \geq 30$, $x_2 \geq 4x_1$, $x_1 \geq 0, x_2 \geq 0$.
- Correct answerDMaximize $z = 1500x_1 + 900x_2$ subject to $x_1 + x_2 \leq 250$, $x_1 \geq 30$, $x_2 \geq 4x_1$, $x_1 \geq 0, x_2 \geq 0$.
Solution
Let $x_1$ be the number of executive class passengers and $x_2$ be the number of economy class passengers.
The profit on each executive class ticket is Rs $1500$ and on each economy class ticket is Rs $900$. The objective is to maximize the total profit $z$, which is given by:
Maximize $z = 1500x_1 + 900x_2$
The airplane can carry a maximum of $250$ passengers, which gives the constraint:
$x_1 + x_2 \leq 250$
The airline reserves at least $30$ seats for executive class, which gives:
$x_1 \geq 30$
At least $4$ times as many passengers prefer to travel by economy class than by executive class, which gives:
$x_2 \geq 4x_1$
Since the number of passengers cannot be negative, we have the non-negativity constraints:
$x_1 \geq 0, x_2 \geq 0$
Combining all these, the Linear Programming Problem is formulated as:
Maximize $z = 1500x_1 + 900x_2$ subject to $x_1 + x_2 \leq 250$, $x_1 \geq 30$, $x_2 \geq 4x_1$, $x_1 \geq 0, x_2 \geq 0$.
Answer: Maximize $z = 1500x_1 + 900x_2$ subject to $x_1 + x_2 \leq 250$, $x_1 \geq 30$, $x_2 \geq 4x_1$, $x_1 \geq 0, x_2 \geq 0$.
The profit on each executive class ticket is Rs $1500$ and on each economy class ticket is Rs $900$. The objective is to maximize the total profit $z$, which is given by:
Maximize $z = 1500x_1 + 900x_2$
The airplane can carry a maximum of $250$ passengers, which gives the constraint:
$x_1 + x_2 \leq 250$
The airline reserves at least $30$ seats for executive class, which gives:
$x_1 \geq 30$
At least $4$ times as many passengers prefer to travel by economy class than by executive class, which gives:
$x_2 \geq 4x_1$
Since the number of passengers cannot be negative, we have the non-negativity constraints:
$x_1 \geq 0, x_2 \geq 0$
Combining all these, the Linear Programming Problem is formulated as:
Maximize $z = 1500x_1 + 900x_2$ subject to $x_1 + x_2 \leq 250$, $x_1 \geq 30$, $x_2 \geq 4x_1$, $x_1 \geq 0, x_2 \geq 0$.
Answer: Maximize $z = 1500x_1 + 900x_2$ subject to $x_1 + x_2 \leq 250$, $x_1 \geq 30$, $x_2 \geq 4x_1$, $x_1 \geq 0, x_2 \geq 0$.
Hence, the correct answer is option: 'D'.