Find the largest trapezoid that can be inscribed in a semicircle

Find the largest trapezoid that can be inscribed in a semicircle with the lower base of the trapezoid on the diameter of the semicircle.

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Let $r$ be the radius of the semicircle, let $b_1$ be the lower edge of the trapezoid, and $b_2$ the upper edge. We recall from geometry that the area of the trapezoid is given by

$A = \frac{1}{2} (b_1 + b_2) h$

where $b_1$ and $b_2$ are the lengths of the bases and $h$ is the height of the trapezoid. Next, we want to find formulas for $b_1$ and $b_2$ in terms of the radius $r$ . Since $b_1$ lies on the diameter of the semicircle we have $b_1 = 2r$ . For $b_2$ we consider the two right triangles drawn in red in the diagram above. The hypotenuse of each of these is $r$ , and the legs have lengths $h$ and $\frac{b_2}{2}$ . Therefore,

$\begin{align*} r^2 = h^2 + \left( \frac{b_2}{2} \right)^2 && \implies && r^2 &= h^2 + \frac{b_2^2}{4} \\[9pt] && \implies && 4r^2 &= 4h^2 + b_2^2 \\[9pt] && \implies && b_2^2 &= 4r^2 - 4h^2 \\[9pt] && \implies && b_2 &= 2 \sqrt{r^2 - h^2}. \end{align*}$

Substituting these into the formula for the area of a trapezoid, we get an expression for the area of the trapezoid in terms of $r$ and $h$ ,

$A = \frac{1}{2} h (b_1 + b_2) = \frac{1}{2} h (2r + \sqrt{2r^2 - h^2}) = hr + h \sqrt{r^2 - h^2}.$

Taking the derivative of this area function with respect to $h$ , (noting that $r$ is a constant since we are given the semicircle of radius $r$ to start with)

$\frac{dA}{dh} = r + \sqrt{r^2 - h^2} - \frac{h^2}{\sqrt{r^2 - h^2}}.$

To find the critical points we set the derivative equal to zero and solve for $h$ ,

$\begin{align*} &&r + \sqrt{r^2 - h^2} - \frac{h^2}{\sqrt{r^2 - h^2}} &= 0 \\ \implies && r \sqrt{r^2 - h^2} + (r^2 - h^2) - h^2 &= 0 \\[9pt] \implies && r \sqrt{r^2 -h^2} &= 2h^2 - r^2 \\[9pt] \implies && r^2 (r^2 - h^2) &= 4h^4 - 4r^2h^2 + r^4 \\[9pt] \implies && r^4 - r^2 h^2 &= 4h^4 - 4r^2h^2 + r^4 \\[9pt] \implies && 4h^4 &= 3r^2 h^2 \\[9pt] \implies && 4h^2 &= 3r^2 \\[9pt] \implies && h &= \frac{\sqrt{3}}{2} r. \end{align*}$

We know the area function has a maximum at this point since for the derivative $\frac{dA}{dh}$ we have

$\begin{align*} \frac{dA}{dh} &> 0 & \text{when} && h &< \frac{\sqrt{3}}{2} r \\ \frac{dA}{dh} &< 0 & \text{when} && h &> \frac{\sqrt{3}}{2} r. \end{align*}$

Hence, $A$ is increasing when $h < \frac{\sqrt{3}}{2} r$ and $A$ is decreasing when $h > \frac{\sqrt{3}}{2} r$ . Thus, the area has a maximum when $h = \frac{\sqrt{3}}{2} r$ .
Finally, we use this value of $h$ to solve for $b_2$ in terms of $r$ ,