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

The area $A$ of the rectangle can be expressed as:

$A = width \times height = (2x)(2y) = 4xy$

From the circle's equation $x^2 + y^2 = 16$, we can solve for $y$:

$y = \sqrt{16 - x^2}$

Substituting $y$ into the area equation gives:

$A = 4x \sqrt{16 - x^2}$

We differentiate the area with respect to $x$:

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

Setting the derivative equal to zero for maximum area:

$16 - 2x^2 = 0 \Rightarrow x^2 = 8 \Rightarrow x = 2\sqrt{2}$

Substituting $x = 2\sqrt{2}$ into the equation for $y$ gives:

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

Therefore, the dimensions of the rectangle are $2x = 4\sqrt{2}$ and $2y = 4\sqrt{2}$ and the maximum area is:

$A = 4xy = 32\text{ square inches}$