Trapezium inscribed in a semicircle

The fourth standard problem: the isosceles trapezium of maximum area inscribed in a semicircle is half of a regular hexagon. A useful trick: maximise $A^2$ to avoid the derivative of the square root.

Example — Isosceles trapezium of maximum area inscribed in a semicircle

An isosceles trapezium has its shorter base varying on the diameter of a semicircle of radius $R$, and its longer base coinciding with the diameter. Find the height that maximises the area.

Solution

Putting $x$ = half the shorter base, the height is $y=\sqrt{R^2-x^2}$ (points on the circumference). Area: $A(x)=\frac{1}{2}(2R+2x)\,y = (R+x)\sqrt{R^2-x^2}, \qquad 0<x<R.$ To avoid the derivative of the square root, I work on $A^2$: $A^2 = (R+x)^2(R^2-x^2) = (R+x)^2(R-x)(R+x) = (R+x)^3(R-x).$ I differentiate with respect to $x$: $(A^2)' = 3(R+x)^2(R-x) - (R+x)^3 = (R+x)^2\bigl[3(R-x)-(R+x)\bigr] = (R+x)^2(2R-4x).$ $(A^2)'=0 \iff x=R/2$ (discarding $x=-R$). Then $y=\sqrt{R^2-R^2/4}=R\sqrt{3}/2$. $A_{\max} = (R+R/2)\cdot R\sqrt{3}/2 = \frac{3R}{2}\cdot\frac{R\sqrt{3}}{2} = \boxed{\frac{3\sqrt{3}}{4}R^2}.$ The maximum trapezium is in fact half of a regular hexagon inscribed in the circle: a geometric surprise.

Links

Topics: Derivatives
Concepts: Objective function · Maximum and minimum
Methods: Maximum minimum via derivative
Skills: Synthetic geometry · Modelling