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Find the largest rectangle that can be inscribed in a semicircle

by RoRi

Given a semicircle of radius r, find the largest rectangle (in terms of volume) that can be inscribed in the semicircle, with base lying on the diameter.

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Let r be the radius of the semicircle, x one half of the base of the rectangle, and y the height of the rectangle. We want to maximize the area, A = 2xy. Referencing the diagram we have

    \[ y = \sqrt{r^2 - x^2}. \]

Thus,

    \[ A = 2x \left(\sqrt{r^2 - x^2}\right) \quad \implies \quad \frac{dA}{dx} = 2 \sqrt{r^2 - x^2} - \frac{2x^2}{\sqrt{r^2 - x^2}}. \]

Setting this derivative equal to 0 and solving for x,

    \begin{align*} \frac{dA}{dx} = 0 && \implies && 2 \sqrt{r^2 - x^2} - \frac{2x^2}{\sqrt{r^2 - x^2}} &= 0 \\ && \implies && 2r^2 - 4x^2 &= 0 \\ && \implies && x = \frac{r}{\sqrt{2}}. \end{align*}

This is a maximum of the area since

    \begin{align*} \frac{dA}{dx} &> 0 & \text{when} && x &< \frac{r}{\sqrt{2}} \\ \frac{dA}{dx} &< 0 & \text{when} && x &> \frac{r}{\sqrt{2}}. \end{align*}

Since y = \sqrt{r^2 - x^2} we then have

    \[ y = \frac{r}{\sqrt{2}}. \]

Thus, the base of the rectangle has length \sqrt{2}r and its height has length \frac{\sqrt{2}}{2} r.

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