More Optimization — Question 2

Question 2 - Solution

Let the top corners of the rectangle be $(x, y)$ and $(-x, y)$, where the rectangle is symmetric about the $y$-axis.

From the parabola: $$y = 12 - x^2$$

The width of the rectangle is $2x$, and the height is $y = 12 - x^2$

Area Function:

$$A(x) = \text{width} \times \text{height} = 2x(12 - x^2) = 24x - 2x^3$$

Differentiate:

$$A'(x) = 24 - 6x^2$$

Set derivative to zero: $$24 - 6x^2 = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2$$

Check Maximum:

Second derivative: $$A''(x) = -12x \Rightarrow A''(2) = -24 < 0$$

So, $x = 2$ gives a maximum.

Final Dimensions:

$$x = 2 \Rightarrow \text{width} = 2x = 4$$ $$y = 12 - x^2 = 12 - 4 = 8$$

$$\boxed{ \begin{aligned} \text{Width} &= 4 \\ \text{Height} &= 8 \\ \text{Maximum Area} &= 32 \end{aligned} }$$