Solutions

Suppose a trapezoid is inscribed in a semicircle of radius $a$, with one base lying along the diameter. We assume that the base spans the entire diameter of the semicircle. If so, then the area is given by

$$\begin{align*} A &= \frac{2a + 2b}{2}\sqrt{a^2 - b^2} \\ &= (a + b)\sqrt{a^2 - b^2}, \end{align*}$$

where $b$ is half the length of the top of the trapezoid. Since $A$ is continuous over $b \in [0, a]$, it must attain its maximum over that interval by the Extreme Value Theorem. The maximum point is in $(0, a)$ because $f(0) = a^2$, $f(a) = 0$ and

$$f\left(\frac{a}{2}\right) = \frac{3\sqrt{3}}{4}a^2 > a^2.$$

The critical point of $A$ is

$$\begin{align*} \frac{dA}{db} = 0 &\implies a\left(\frac{1}{2}\right)(a^2 - b^2)^{-1/2}(-2b) + (a^2 - b^2)^{1/2} + b\left(\frac{1}{2}\right)(a^2 - b^2)^{-1/2}(-2b) = 0 \\ &\implies \frac{-2b(a+b)}{2\sqrt{a^2 - b^2}} + \sqrt{a^2 - b^2} = 0 \\ &\implies -b(a+b) + a^2 - b^2 = 0 \\ &\implies -2b^2 - ab + a^2 = 0 \\ &\implies -2b^2 - 2ab + ab + a^2 = 0 \\ &\implies -2b(b+a) + a(b+a) = 0 \\ &\implies (a - 2b)(b+a) = 0 \\ &\implies b = \frac{a}{2}, \end{align*}$$

since $a = -b$ is inadmissible, and this is the maximum point.