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Question #41760

Difficulty: 8/10 | 🟢 Free

The Question

Let $Z(x,y)$ be the objective function defined by the line integral: $Z(x,y) = \int_{(0,0)}^{(x,y)} \left( (v e^{u^2 v^2} + 2u)du + (u e^{u^2 v^2} + 2v)dv \right)$ where the integral is taken over any piecewise smooth curve connecting $(0,0)$ to $(x,y)$ . Consider the linear programming problem: Maximize $Z(x,y)$ subject to the constraints: $2x + y \ge 4$ $x - y \le 1$ $x \ge 0, y \ge 0$ Which of the following represents the maximum value of $Z(x,y)$ ?

Options

A.
16
B.
$\int_{0}^{10/9} e^{t^2} dt + \frac{29}{9}$
C.
20
D.
\text{Does not exist}

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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

Boundary Value Theorem|Exact Differential Recognition|Unbounded Region Trap|Inequalities in Definite Integrals

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to integrate Exact Differential Recognition with Linear Programming, specifically targeting the 'Unbounded Region Trap'.

Students often blindly apply the Boundary Value Theorem, evaluating the objective function only at the vertices $(0,4)$ and $(5/3, 2/3)$ to incorrectly conclude the maximum is 16.

However, the feasible region is unbounded.

By recognizing the exact differential, the objective function simplifies to $Z(x,y) = \int_{0}^{xy} e^{t^2} dt + x^2 + y^2$ .

Applying integral inequalities ( $e^{t^2} \ge 1$ ), one can prove that along the asymptote $y=x$ , $Z(x,x) \ge 3x^2 \to \infty$ as $x \to \infty$ .

Thus, the vertex approach is conceptually hollow here, and the maximum does not exist.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Programming - Class 12
Concept	Boundary Value Theorem
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free