Maximum volume of a cone

If the radius of the base is $r$ and the height is $h$ the volume of the cone is $\dfrac{1}{3}\pi r^2h$ and the surface area (excluding the base) is $\pi r\sqrt{r^2+h^2}$ .

From the hypothesis, we have $\pi r\sqrt{r^2+h^2}=100$ . And we must find the maximum value of $\dfrac{1}{3}\pi r^2h$ .

Appying the AM-GM inequality, we have: $r^2+h^2=r^2+\dfrac{h^2}{2}+\dfrac{h^2}{2}\ge3\sqrt[3]{\dfrac{r^2h^4}{4}}$ Hence, $100=\pi r\sqrt{r^2+h^2}\ge\pi r\cdot \sqrt3\sqrt[3]{\dfrac{rh^2}{2}}=\pi\sqrt3\sqrt[3]{\dfrac{r^4h^2}{2}}$

This implies that $r^4h^2\le2\left(\dfrac{100}{\pi\sqrt3}\right)^3=\dfrac{2\cdot 10^6}{3\pi^3\sqrt3}$ or $r^2h\le\dfrac{1000\sqrt2}{\pi\sqrt{\pi}\cdot\sqrt[4]{27}}$

So, $\dfrac{1}{3}\pi r^2h\le\dfrac{1000\sqrt2}{3\sqrt{\pi}\cdot\sqrt[4]{27}}\approx\boxed{116.675}$ .

The inequality holds if and only if $h=r\sqrt2=\dfrac{10\sqrt2}{\sqrt{\pi}\cdot\sqrt[4]{3}}$ .

For maximum volume of a right circular cone (open on top), $h=r\sqrt{2}$ .

$L=\sqrt{h^2+r^2}=\sqrt{2r^2+r^2}=r\sqrt{3}$

The lateral area of the cone is given by $A=\dfrac{1}{2}cL$ where $c$ is the circumference of the base and $L$ is the slant height. We have

$100=\dfrac{1}{2}(2\pi r)(r\sqrt{3})$

$r^2=\dfrac{100}{\sqrt{3}\pi}$ or $r=\sqrt{\dfrac{100}{\sqrt{3}\pi}}=\dfrac{10}{\sqrt{\sqrt{3}\pi}}$

The volume of a cone is given by $V=\dfrac{1}{3}A_b h$ where $A_b$ is the area of the base and $h$ is the height. We have

$V=\dfrac{1}{3}(\pi)\left(\dfrac{100}{\sqrt{3}\pi}\right)\left(\dfrac{10}{\sqrt{\sqrt{3}\pi}}\sqrt{2}\right) \approx \boxed{116.675}$