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Question #41671

Difficulty: 8/10 | 🟢 Free

The Question

Let $S = \left\{ (x, y) \in \mathbb{R}^2 : 0 \le x \le 5, 0 \le y \le 5, \frac{x - 3}{y - 2} \ge 1 \right\}$ . Find the supremum of the function $f(x, y) = x^2 + y^2 - 10x - 10y + 50$ over the set $S$ .

Options

A.
13
B.
25
C.
41
D.
50

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

Feasible Region Constancy|Maximum/Minimum of Quadratic Expressions|Cross Multiplication Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Cross Multiplication Trap' in algebraic inequalities combined with quadratic optimization over a polyhedral region.

Students often blindly cross-multiply the constraint $\frac{x-3}{y-2} \ge 1$ to get $x - 3 \ge y - 2 \implies y \le x - 1$ , completely ignoring the case where the denominator $y - 2 < 0$ flips the inequality.

If they fall for this trap, they map a false feasible region (a triangle with vertices $(1,0), (5,0), (5,4)$ ) and incorrectly find the maximum squared distance to be 41 at $(1,0)$ .

By properly analyzing the sign of the denominator, the true feasible region includes a second component containing the origin $(0,0)$ , which yields the correct global supremum of 50.

Quick Facts About This Question

Question Information

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