A company is designing a logo - AQA - A-Level Maths Pure - Question 13 - 2018 - Paper 1

The area $A$ of the rectangle can be calculated using the formula:

$A = ext{width} imes ext{height} = 2x imes 2y = 4xy$

From the equation of the circle, solve for $y$:

$y = ext{sqrt}(16 - x^2)$

Thus, we can substitute $y$ into the area equation:

$A = 4x ext{sqrt}(16 - x^2)$

To find the maximum area, differentiate $A$ with respect to $x$:

$\frac{dA}{dx} = 4\left( \text{sqrt}(16 - x^2) + x\left( \frac{-x}{\text{sqrt}(16 - x^2)} \right) \right)$

Set the derivative equal to zero for critical points.

Set the derivative $\frac{dA}{dx} = 0$ and solve for $x$:

$4\left( \text{sqrt}(16 - x^2) - \frac{x^2}{\text{sqrt}(16 - x^2)} \right) = 0$

From this, we find that $x = \sqrt{8}$.

To confirm that this value gives a maximum area, we use the second derivative test or evaluate the area function at values around $x = \sqrt{8}$.

Calculate the area when $x = 2.8$ and $x = 2.9$ to confirm maximum.

Substituting $x = 2$ into the area function gives:

$A = 4 \cdot 2 \cdot \text{sqrt}(16 - 2^2) = 32 \text{ square inches}.$

Therefore, the maximum area is 32 square inches.