Q.15. A rectangle is drawn in the region enclosed by the curves $p(x) = 8 - 2x^2$ and $q(x) = x^2 - 2$.

The sides of the rectangle are parallel to the x- and y-axes.

What is the maximum possible area of the rectangle?

Answer Choices

• $\frac{26}{9}$
• $\frac{52}{9}$
• $\frac{4\sqrt{6}}{3}$
• $\frac{8\sqrt{6}}{3}$
• $4\sqrt{2}$
• $8\sqrt{2}$
• $\frac{20\sqrt{10}}{9}$
• $\frac{40\sqrt{10}}{9}$

Solution

The area of the rectangle is given by $A(x) = 2x(10 - 3x^2)$. Maximizing this function gives a critical point at $x = \frac{\sqrt{10}}{3}$. Substituting this value into the area function yields the maximum possible area of:

$$A = \frac{40\sqrt{10}}{9}$$

The correct answer is: $\frac{40\sqrt{10}}{9}$ .