Rectangle inscribed in a semicircle

Statement

A rectangle is inscribed in a semicircle of radius $R$ (with one side on the diameter). Which is the rectangle of maximum area?

Solution

Let $x$ be the half-base (half of the side on the diameter), with $0<x<R$; the height reaches the circumference, so $y=\sqrt{R^2-x^2}$. Area: $A(x)=2x\,y = 2x\sqrt{R^2-x^2}.$ To avoid the root I maximise $A^2 = 4x^2(R^2-x^2)=4(x^2R^2-x^4)$: $(A^2)' = 4(2xR^2-4x^3)=8x(R^2-2x^2)=0 \iff x=\frac{R}{\sqrt2}.$ Then $y=\sqrt{R^2-R^2/2}=\dfrac{R}{\sqrt2}$: the optimal rectangle has base $2x=R\sqrt2$ and height $R/\sqrt2$. $A_{\max}=2\cdot\frac{R}{\sqrt2}\cdot\frac{R}{\sqrt2}=\boxed{R^2}.$

Links

Topics: Derivatives
Concepts: Objective function · Maximum minimum
Methods: Maximum minimum via derivative
Skills: Synthetic geometry · Modelling
Exercise type: Geometric problem