MHT CET · Maths · Linear Programming

The minimum value of the objective function \(z=4 x+6 y\) subject to \(x+2 y \geq 80,3 x+y \geq 75, x, y \geq 0\) is

A 324
B 250
C 320
D 254

Verified Solution

Answer & Solution

Correct Answer

(D) 254

Step-by-step Solution

Detailed explanation

Refer figure
Required area is shaded. Vertices of feasible region are \(\mathrm{A}=(80,0) ; \mathrm{C}=(0,75)\) and point of intersection of given lines is \(\mathrm{B}=(14,33)\)
We have to minimize objective function \(\mathrm{z}=4 \mathrm{x}+6 \mathrm{y}\)
\(
\begin{aligned}
& \therefore \quad \mathrm{Z}_{(\mathrm{A})}=4(80)+6(0)=320 \\
& \mathrm{Z}_{(\mathrm{B})}=4(14)+6(33)=254 \\
& \mathrm{Z}_{(\mathrm{C})}=4(0)+6(75)=450 \\
&
\end{aligned}
\)

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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