KCET · Maths · Linear Programming

The shaded region is the solution set of the inequalities

A \(5 x+4 y \geq 20, x \leq 6, y \geq 3, x \geq 0, y \geq 0\)
B \(5 x+4 y \leq 20, x \leq 6, y \leq 3\), \(x \geq 0, y \geq 0\)
C \(5 x+4 y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0\)
D \(5 x+4 y \geq 20, x \geq 6, y \leq 3, x \geq 0, y \geq 0\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(5 x+4 y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0\)

Step-by-step Solution

Detailed explanation

Lets draw the given shaded region,

Line \(l_{1} \Rightarrow \frac{x}{4}+\frac{y}{5}=1 \quad\) (intercept form) \(\Rightarrow \quad \frac{5 x+4 y}{4 \times 5}=1\) \(\Rightarrow \quad 5 x+4 y=20\) As, origin is not in the feasible region. \(\therefore \quad 5 x+4 y \geq 20\) Line \(l_{2} \Rightarrow y \leq 3\) (from the graph) Line \(l_{3} \Rightarrow x \leq 6\) (from the graph) and coordinate axes \(x \geq 0, y \geq 0\) Hence, inequalities are \(5 x+4 y \geq 20, y \leq 3, x \leq 6\), \(x \geq 0, y \geq 0 .\)

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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