Cylinder in a Cone

Base from the figure above, we express $h$ in terms of $r$ by ratio and proportion,

$\frac{5-r}{h}=$ $\frac{5}{12}$

$h=$ $\frac{12}{5}$ $(5-r)$

We know that the volume of a right circular cylinder is

$V =$ $area$ $of$ $the$ $base$ $x$ $height=$ $πr^2h=$ $πr^2($ $\frac{12}{5})(5-r)=$ $12πr^2-$ $\frac{12}{5}r^3$

Now differentiate both sides with respect to $r$ ,

$\frac{dV}{dr}=24πr-$ $\frac{36}{5}πr^2$

set $\frac{dV}{dr}$ to $0$

$24πr=$ $\frac{36}{5}πr^2$

$r=$ $\frac{5(24)}{36}=$ $\frac{10}{3}$ $cm$

We need to show that $V$ is maximized when $r=$ $\frac{10}{3}$ , applying the second derivative test shows that $\frac{d^2V}{dr^2}=$ $24π -$ $\frac{72}{5}$ $πr<0$ . This concludes that $V$ is indeed maximized when $r=$ $\frac{10}{3}$ .

Now, solving for $h$ , we have

$h=$ $\frac{12}{5}(5-$ $\frac{10}{3})=4$ $cm$