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Question #41761

Difficulty: 8/10 | 🟢 Free

The Question

Let $R$ be the feasible region in the first quadrant ( $x \ge 0, y \ge 0$ ) satisfying the following constraints:

$\left( \int_{-\pi/4}^{\pi/4} \ln\left(t + \sqrt{1+t^2}\right) dt \right) x + x + y \le 4$
$x \sin \alpha + y \cos \alpha \le 5$ for all $\alpha \in [0, 2\pi]$

Consider the objective function $Z(x,y) = \int_{0}^{x^2 y} \frac{4-t}{1+t^2} dt$ . The maximum value of $Z(x,y)$ over the region $R$ is:

Options

A.
$4 \arctan(4) - \frac{1}{2} \ln(17)$
B.
$0$
C.
$4 \arctan\left(\frac{256}{27}\right) - \frac{1}{2} \ln\left(1 + \left(\frac{256}{27}\right)^2\right)$
D.
$4 \arctan(8) - \frac{1}{2} \ln(65)$

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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|Fundamental Theorem of Calculus|Even and Odd Function Symmetry|Auxiliary Angle Method

Why This Question Is Tricky

Common Mistakes

The question fuses Linear Programming with Calculus to test the Fundamental Theorem of Calculus (FTC) under optimization constraints.

Students typically evaluate LPP objective functions only at the boundary vertices (e.g., $(4,0)$ or $(0,4)$ ), which yields $Z=0$ .

Alternatively, they might maximize the upper bound $u = x^2y$ to $256/27$ and incorrectly plug this into the integral.

The trap lies in recognizing that the objective function is a composite integral function $Z(u)$ .

By applying the Leibniz rule, the true maximum occurs at the critical point $u=4$ inside the integral's domain, provided $u=4$ is reachable within the feasible region.

The constraints themselves are masked using odd-function symmetry and the auxiliary angle method to increase cognitive load.

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