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Question #41769

Difficulty: 8/10 | 🟢 Free

The Question

Let $R$ be the feasible region in the first quadrant ( $x \ge 0, y \ge 0$ ) defined by the inequalities: $\sqrt{x^2 + (y - 60\sqrt{2})^2} \le \sqrt{2} (y - 30\sqrt{2})$ $\left\{ \frac{x}{32} \right\} \le \frac{11}{32}$ where $\{ \cdot \}$ denotes the fractional part function. Find the minimum value of the objective function $Z = 13y - 5x$ over the region $R$ .

Options

A.
720
B.
724
C.
738
D.
\text{Unbounded}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Unbounded Region Trap|General Conic Definition|Symmetry and Periodicity

Why This Question Is Tricky

Common Mistakes

This question fuses the focal definition of a hyperbola with a periodic LPP constraint to create a sophisticated 'Unbounded Region Trap'.

Students often ignore the periodic constraint and find the unconstrained minimum (720).

If they do apply the constraint, they typically only check the first feasible interval $[0, 11]$ and incorrectly assume the local minimum at $x=11$ (738) is the global minimum.

They fail to realize that the objective function's convexity makes the boundary of the second interval $x=32$ (724) the true global minimum.

Additionally, the negative coefficient in the objective function tempts students to incorrectly guess 'Unbounded' without performing the asymptotic analysis.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Programming - Class 12
Concept	Unbounded Region Trap
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free