Find all local extreme values of V² on 0<x<1.

01 / Standard#Your turn

V(x)=\frac\pi3x^2\sqrt{1-x^2},\quad0\le x\le1

Official paper · jm02-2024 · 2(a)(ii) · PDF 4

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Differentiate the polynomial V², not the square-root volume.

Hint 2

Check derivative signs at the interior stationary point.

Worked solution

The only interior stationary point is √(2/3).
$(V^2)'=\frac{2\pi^2}{9}x^3(2-3x^2)=0\implies x=\frac{\sqrt6}3$
Its derivative changes from positive to negative. There is no interior local minimum.
$V^2(\sqrt6/3)=\frac{\pi^2}9\left(\frac49-\frac8{27}\right)=\frac{4\pi^2}{243}$

Local maximum 4π²/243 at x=√6/3; no local minimum on (0,1).

Checks and common pitfalls: Do not count endpoint minima in an explicitly open interval.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The only interior stationary point is √(2/3).
$(V^2)'=\frac{2\pi^2}{9}x^3(2-3x^2)=0\implies x=\frac{\sqrt6}3$
Its derivative changes from positive to negative. There is no interior local minimum.
$V^2(\sqrt6/3)=\frac{\pi^2}9\left(\frac49-\frac8{27}\right)=\frac{4\pi^2}{243}$

Think first. Reveal a hint when the class is ready.

2024 JM02