Sample 2024-25 Class 12 Maths Q11 Solution: Where Z = 4x + 3y is maximum on a bounded feasible region

The question

For the linear programming problem (LPP), the objective function is $Z = 4x + 3y$ and the feasible region determined by a set of constraints is shown in the graph:

The given graph, redrawn.

(Note: The figure is not to scale.)

Which of the following statements is true?

(A) Maximum value of $Z$ is at $R(40, 0)$ ⁠.

(B) Maximum value of $Z$ is at $Q(30, 20)$ ⁠.

(C) Value of $Z$ at $R(40, 0)$ is less than the value at $P(0, 40)$ ⁠.

(D) The value of $Z$ at $Q(30, 20)$ is less than the value at $R(40, 0)$ ⁠.

Exam answer

The corner points of the feasible region are $O$ ⁠, $R$ ⁠, $Q$ and $P$ ⁠.

Values of $Z = 4x + 3y$ at the corner points:

\begin{aligned} O(0, 0)&: Z = 0 \\ R(40, 0)&: Z = 160 \\ Q(30, 20)&: Z = 120 + 60 = 180 \\ P(0, 40)&: Z = 120 \end{aligned}

Why?

Put each corner into $Z = 4x + 3y$ ⁠. For example, at $Q$ ⁠: $4(30) = 120$ and $3(20) = 60$ ⁠. At $R$ ⁠: $4(40) = 160$ ⁠. At $P$ ⁠: $3(40) = 120$ ⁠.

\begin{aligned} O&: 4(0) + 3(0) = 0 + 0 = 0 \\ R&: 4(40) + 3(0) = 160 + 0 \\ &\quad = 160 \\ Q&: 4(30) + 3(20) = 120 + 60 \\ &\quad = 180 \\ P&: 4(0) + 3(40) = 0 + 120 \\ &\quad = 120 \end{aligned}

The feasible region is bounded, so the maximum value of $Z$ is 180, at $Q(30, 20)$ ⁠.

Why?

Corner point rule: on a bounded feasible region, $Z$ has a maximum and a minimum, and each occurs at a corner point. So the largest of the four corner values is the maximum.

Watch out: $R$ has the largest $x$ ⁠, but $Q$ gives the largest $Z$ ⁠, because $y$ counts too.

So (A) is false. Also $Z(R) = 160 > 120 = Z(P)$ ⁠, so (C) is false, and $Z(Q) = 180 > 160 = Z(R)$ ⁠, so (D) is false.

Answer: (B) Maximum value of $Z$ is at $Q(30, 20)$ ⁠.

Where this comes from

Chapter 12, Linear Programming: LPP: maximum at a corner point.

Question type: Find the maximum

How to spot it: click to see

Type 1. Find the maximum: lines, shading, corners, table

Spot it: "Maximise $Z$ " with " $\le$ " rules that have only plus signs, like $x + 2y \le 10$ ⁠. The region is then a closed shape with a corner at the origin.

Here: The region OPQR has a corner at the origin and is bounded, so put the four corners into $Z = 4x + 3y$ ⁠. The largest value, 180 at $Q(30, 20)$ ⁠, is the maximum.

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Q11: Where Z=4x+3yZ = 4x + 3yZ=4x+3y is maximum on a bounded feasible region

Q11: Where $Z = 4x + 3y$ is maximum on a bounded feasible region