Answer to Question #168984 in Calculus for Vishal

Question

A right triangle with hypotenuse of length 'a' is rotated about one of its legs to generate a right circular cone. Find the greatest possible volume of such a cone.

Answer 1

Volume of the cone

$V = \frac13 \pi r^2h$ /1/

where

r - radius of the base

h - height

In this problem 'r' and 'h' are the legs of the right triangle with hypotenuse 'a', so

$r^2 + h^2 = a^2$ /2/

Express the 'r' in terms of the 'h' using /2/ and substitute it into the expression for volume /1/

$r^2 = a^2 - h^2$

$V = \frac13 \pi (a^2 - h^2)h = \frac13 \pi(a^2h - h^3)$ /3/

Find derivative of the volume (assumes that the 'a' is meant to be constant)

$V' = \frac13\pi(a^2 - 3h^2)$

Equate the derivative to zero and calculate 'h'

$\frac13\pi(a^2 - 3h^2) = 0$

$h = \cfrac{a}{\sqrt3}$ /4/

Find the second derivative for check is volume has maximum with given 'h'

$V'' = \frac13 \pi(-6h) = -2\pi h$

$V'' < 0$ for any positive 'h', so volume has a maximum

Plug obtained 'h' from /4/ into /3/ and find the maximum volume

$V_{max} = \frac13\pi(a^2 - (\frac{a}{\sqrt3})^2)\frac{a}{\sqrt3} = \frac{2}{9\sqrt3}\pi\,a^3$

Answer: greatest possible volume is $\frac{2}{9\sqrt3}\,\pi\,a^3$