MHT CET · Maths · Linear Programming
If feasible region is as shown in the figure, then the related inequalities are

- A \(3 x+4 y \geq 12, y-x \geq 0, y \leq 3, x, y \geq 0\)
- B \(3 x+4 y \leq 12, y-x \leq 0, y \geq 3, x, y \geq 0\)
- C \(3 x+4 y \leq 12, y-x \geq 0, y \leq 3, x, y \geq 0\)
- D \(3 x+4 y \geq 12, y-x \leq 0, y \geq 3, x, y \geq 0\)
Answer & Solution
Correct Answer
(A) \(3 x+4 y \geq 12, y-x \geq 0, y \leq 3, x, y \geq 0\)
Step-by-step Solution
Detailed explanation
The shaded region lies:
on non-origin side of line \(3 x+4 y=12\) i.e., \(3 x+4 y \geq 12\),
on the side of the line \(y-x=0\), where \(y \geq x\) i.e., \(y-x \geq 0\),
on origin side of line \(y=3\)
i.e., \(y \leq 3\),
and in first quadrant
i.e., \(x \geq 0, y \geq 0\).
on non-origin side of line \(3 x+4 y=12\) i.e., \(3 x+4 y \geq 12\),
on the side of the line \(y-x=0\), where \(y \geq x\) i.e., \(y-x \geq 0\),
on origin side of line \(y=3\)
i.e., \(y \leq 3\),
and in first quadrant
i.e., \(x \geq 0, y \geq 0\).
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