COMEDK · Maths · 35. Linear Programming

The minimum value of \(Z=150 x+200 y\) for the given constraints
\(\begin{aligned}
& 3 x+5 y \geq 30 \\
& x+y \geq 8 ; x \geq 0, y \geq 0 \text { is }
\end{aligned}\)

A 1350
B 1600
C 1200
D 0

Verified Solution

Answer & Solution

Correct Answer

(A) 1350

Step-by-step Solution

Detailed explanation

Corner points of the feasible region are found by intersecting the boundary lines. Intersection of \(3x + 5y = 30\) and \(x + y = 8\): Subtracting \(3(x+y=8)\) from first: \(2y = 6 \Rightarrow y = 3,\ x = 5\). Point: \((5, 3)\) Other corner points: \((10, 0)\) and \((0, 8)\).…

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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The minimum value of \(Z=150 x+200 y\) for the given constraints\(\begin{aligned}& 3 x+5 y \geq 30 \\& x+y \geq 8 ; x \geq 0, y \geq 0 \text { is }\end{aligned}\)

Step-by-step Solution

The minimum value of \(Z=150 x+200 y\) for the given constraints
\(\begin{aligned}
& 3 x+5 y \geq 30 \\
& x+y \geq 8 ; x \geq 0, y \geq 0 \text { is }
\end{aligned}\)