a rectangle is constructed with its base on the diameter of a semi - circle with radius 22 and with its two…

Explanation:

Step1: Differentiate the area function

Let $A(x)=x\sqrt{484 - \frac{x^{2}}{4}}$. First, rewrite it as $A(x)=x(484-\frac{x^{2}}{4})^{\frac{1}{2}}$. Using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x$ and $v=(484-\frac{x^{2}}{4})^{\frac{1}{2}}$. $u^\prime=1$ and $v^\prime=\frac{1}{2}(484 - \frac{x^{2}}{4})^{-\frac{1}{2}}\times(-\frac{x}{2})$. Then $A^\prime(x)=\sqrt{484-\frac{x^{2}}{4}}+x\times\frac{1}{2}(484 - \frac{x^{2}}{4})^{-\frac{1}{2}}\times(-\frac{x}{2})=\sqrt{484-\frac{x^{2}}{4}}-\frac{x^{2}}{4\sqrt{484-\frac{x^{2}}{4}}}=\frac{484-\frac{x^{2}}{4}-\frac{x^{2}}{4}}{\sqrt{484-\frac{x^{2}}{4}}}=\frac{484 - \frac{x^{2}}{2}}{\sqrt{484-\frac{x^{2}}{4}}}$.

Step2: Find the critical points

Set $A^\prime(x) = 0$. Then $484-\frac{x^{2}}{2}=0$ (since the denominator $\sqrt{484-\frac{x^{2}}{4}}\neq0$ for the domain where the function is well - defined). Solving $484-\frac{x^{2}}{2}=0$ gives $\frac{x^{2}}{2}=484$, so $x^{2}=968$ and $x = 22\sqrt{2}$ (we take the positive value since $x\in[0,44]$).

Step3: Find the height

The equation of the semi - circle is $y=\sqrt{484 - \frac{x^{2}}{4}}$. Substitute $x = 22\sqrt{2}$ into the equation of the semi - circle to find the height. $y=\sqrt{484-\frac{(22\sqrt{2})^{2}}{4}}=\sqrt{484 - 242}=\sqrt{242}=11\sqrt{2}$. The base of the rectangle is $x = 22\sqrt{2}$ and the diameter of the semi - circle is used for the base of the rectangle, so the base of the rectangle is $2x=44\sqrt{2}$.

Answer:

Base: $44\sqrt{2}$, Height: $11\sqrt{2}$