Sample 2023-24 Class 12 Maths Q7 Solution: Condition on p, q for the minimum of Z = px + qy at (3, 0) and (1, 1)

The question

The corner points of the bounded feasible region determined by a system of linear constraints are $(0, 3)$ ⁠, $(1, 1)$ and $(3, 0)$ ⁠. Let $Z = px + qy$ ⁠, where $p, q > 0$ ⁠. The condition on $p$ and $q$ so that the minimum of $Z$ occurs at $(3, 0)$ and $(1, 1)$ is

(a) $p = 2q$

(b) $p = \frac{q}{2}$

(c) $p = 3q$

(d) $p = q$

Exam answer

\begin{aligned} \text{At } (3, 0)&: Z = 3p \\ \text{At } (1, 1)&: Z = p + q \end{aligned}

Why?

For a bounded feasible region, the minimum of $Z$ occurs at a corner point. So we work out $Z = px + qy$ at the corners.

At $(3, 0)$ ⁠: $Z = p(3) + q(0) = 3p$ ⁠. At $(1, 1)$ ⁠: $Z = p(1) + q(1)$ ⁠.

For the minimum at both points, $3p = p + q$ ⁠, so $q = 2p$ ⁠.

Why?

If the minimum occurs at two corners, $Z$ takes the same smallest value at both. So the two values are equal, and $3p - p = q$ gives $2p = q$ ⁠.

\begin{aligned} 3p &= p + q \\ 3p - p &= q \quad (\text{take } p \text{ from both sides}) \\ 2p &= q \\ p &= \frac{q}{2} \quad (\text{divide both sides by } 2) \end{aligned}

$\therefore\ p = \frac{q}{2}$

Then at $(0, 3)$ ⁠, $Z = 3q = 6p > 3p$ ⁠, so the minimum is indeed at $(3, 0)$ and $(1, 1)$ ⁠.

Why?

The third corner must give a larger value, or it would share the minimum. With $q = 2p$ ⁠, $Z(0, 3) = 3q = 6p$ ⁠, and $6p > 3p$ because $p > 0$ ⁠.

Then $Z = p(x + 2y)$ ⁠, which is the same at every point of the edge joining $(3, 0)$ and $(1, 1)$ ⁠: that edge lies on $x + 2y = 3$ ⁠.

Answer: (b) $p = \dfrac{q}{2}$

Where this comes from

Chapter 12, Linear Programming: Corner point method: minimum at two corners.

Question type: Both maximum and minimum, and a tie along an edge

How to spot it: click to see

Type 3. Both maximum and minimum, and a tie along an edge

Spot it: "Minimise and maximise $Z$ ". Often one rule, like $x - 2y \ge 0$ ⁠, is a line through the origin.

Here: A minimum at both $(3, 0)$ and $(1, 1)$ means the two corner values tie: $3p = p + q$ ⁠, so $q = 2p$ ⁠, that is $p = \frac{q}{2}$ ⁠.

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Q7: Condition on p,qp, qp,q for the minimum of Z=px+qyZ = px + qyZ=px+qy at (3,0)(3, 0)(3,0) and (1,1)(1, 1)(1,1)

Q7: Condition on $p, q$ for the minimum of $Z = px + qy$ at $(3, 0)$ and $(1, 1)$