Answer to Question #229571 in Calculus for Natty Des

Question

3. Find the maximum volume of a right - circular cylinder that can be inscribed in a right-circular cone of
radius R = 3 and height H = 9.

Answer 1

Let radius of cylinder be r and height be h as in the diagram above.

So in ∆ABC and ∆ADE;

$\frac{AB}{AD}=\frac{BC}{DE}$

=$\frac{(H-h)}{H}=\frac{r}{R}$

$\implies r=\frac{(H-h)}{H}R$

Volume of cylinder=πr²h

=$π*(\frac{(H-h)^{2}}{H^{2}})R^{2}*h$

V(h)=$\frac{πR^{2}}{H^{2}}(H^{2}+h^{2}-2Hh)h$

=$\frac{πR^{2}}{H^{2}}*(H^{2}h+h^{3}-2h^{2}H)$

$\frac{dV(h)}{dh}=0$ for maximum volume

So we differentiate (H²h+h³-2h²H) with respect to h

=H²+3h²-4hH=0

3h²-3hH-hH+H²=0

3h(h-H)-H(h-H)=0

(h-H)(3h-H)=0

$\therefore$ h=$\frac{H}{3}$

V=$\frac{πR^{2}}{H^{2}}*[(H^{2}*\frac{H}{3})+(\frac{H}{3})^{3}-2(\frac{H}{3})^{2}*H]$

=$\frac{πR^{2}}{H^{2}}*[\frac{H^{3}}{3}+\frac{H^{3}}{27}-\frac{2H^{3}}{9}]$

=$\frac{πR^{2}}{H^{2}}*\frac{4H^{3}}{27}$

=4πH$\frac{R^{2}}{27}$

V_max=$4πH\frac{R^{2}}{27}$

r=3, H=9

V_max=$4*π*9*\frac{3^{2}}{27}$

=37.699 cubic units