Rectangle of maximum area in a semicircle

Statement

A rectangle with one side on the diameter is inscribed in a semicircle of radius $r$. Find the dimensions that maximise the area.

Rectangle inscribed in the semicircle: base $2x$, height $y$.

Solution

With $x^2+y^2=r^2$ and area $A=2xy$, substituting $y=\sqrt{r^2-x^2}$: $A(x)=2x\sqrt{r^2-x^2}.$ Squaring: $A^2=4x^2(r^2-x^2)=-4x^4+4r^2x^2.$ It is a parabola in the variable $x^2$, with downward concavity: the maximum is at the vertex, that is at $x^2=\dfrac{r^2}{2}$, from which $x=\dfrac{r}{\sqrt2}$ and $y=\sqrt{r^2-\dfrac{r^2}{2}}=\dfrac{r}{\sqrt2}$.

Dimensions: base $=2x=r\sqrt2$, height $=y=\dfrac{r}{\sqrt2}=\dfrac{r\sqrt2}{2}$, with maximum area $A_{\max}=\boxed{r^2}$.

Links

Topics: Parabola
Concepts: Parabola · Vertex
Functions: Parabola
Skills: Analytic geometry · Modelling
Exercise type: Geometric problem