Q80 / 150 · MHT CET 2026 (11 April Shift 2) · Mathematics
The corner points of the feasible region determined by a system of linear constraints are $(0, 3), (1, 1)$ and $(3, 0)$. If the objective function is $z = px + qy$, where $p, q > 0$, then the condition on p and q such that the minimum of z occurs at both $(3, 0)$ and $(1, 1)$ is _____
- A$p = 3q$
- B$3p = q$
- Correct answerC$p = \dfrac{q}{2}$
- D$p = 2q$
Solution
The objective function is $z = px + qy$.
The value of $z$ at the corner point $(3, 0)$ is $z = 3p + 0 = 3p$.
The value of $z$ at the corner point $(1, 1)$ is $z = p + q$.
Since the minimum of $z$ occurs at both $(3, 0)$ and $(1, 1)$, the values of $z$ at these points must be equal.
$3p = p + q$
$2p = q$
$p = \dfrac{q}{2}$
Checking the value at $(0, 3)$, $z = 3q = 6p$. Since $p > 0$, $3p < 6p$, confirming the minimum is indeed at $(3, 0)$ and $(1, 1)$.
Answer: $p = \dfrac{q}{2}$
The value of $z$ at the corner point $(3, 0)$ is $z = 3p + 0 = 3p$.
The value of $z$ at the corner point $(1, 1)$ is $z = p + q$.
Since the minimum of $z$ occurs at both $(3, 0)$ and $(1, 1)$, the values of $z$ at these points must be equal.
$3p = p + q$
$2p = q$
$p = \dfrac{q}{2}$
Checking the value at $(0, 3)$, $z = 3q = 6p$. Since $p > 0$, $3p < 6p$, confirming the minimum is indeed at $(3, 0)$ and $(1, 1)$.
Answer: $p = \dfrac{q}{2}$
Hence, the correct answer is option: 'C'.