The maximum value of z = 6x +8y subject to x-y≥0,x+3y≤12,x≥0,y≥0 is…..

MHT CET · Maths · Straight Lines

The maximum value of z = 6x +8y subject to $x - y \geq 0, x + 3 y \leq 12, x \geq 0, y \geq 0$ is…..

A 72
B 42
C 96
D 24

Verified Solution

Answer & Solution

Correct Answer

(A) 72

Step-by-step Solution

Detailed explanation

We have, z = 6x +8y
Subject to constraints

x - y \geq 0, x + 3 y \leq 12, x \geq 0, y \geq 0

.
On taking given constraints as equations, we get the following graph

Intersecting point of the line

x - y = 0

and

x + 3 y = 12

is B (3, 3).
Here, OABO is the required feasible region
Whose corner points are O (0, 0), A (12, 0) and B (0, 4) Now,
Corner points

Z = 6 x + 8 y

O (0, 0)

6 \times 0 + 8 \times 0 = 0

A (12, 0)

6 \times 12 + 8 \times 0 = 72

(maximum)
B (3, 3)

6 \times 3 + 8 \times 3 = 42

∴

Maximum value of Z is 72.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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