Maximum Volume

Computation of the volume of the cone so that it is maximum

By pythagorean theorem, we have

$r^2=6^2-(h-6)^2=36-(h^2-12h+36)=36-h^2+12h-36=-h^2+12h$

The volume of a right circular cone is given by the formula,

$V=\dfrac{1}{3}\pi r^2h$

Substituting, we get

$V=\dfrac{1}{3}\pi (-h^2+12h)(h)=\dfrac{1}{3}\pi (-h^3+12h^2)$

Differentiating with respect to $h$ , gives us

$\dfrac{dV}{dh}=\dfrac{1}{3}\pi (-3h^2+24h)$

$\dfrac{dV}{dh}=0$

$3h^2=24h$

$h=8$

It follows that,

$r=\sqrt{6^2-(8-6)^2}=\sqrt{36-4}=\sqrt{32}=\sqrt{16(2)}=4\sqrt{2}$

Then,

$L=\sqrt{r^2+h^2}=\sqrt{(4\sqrt{2})^2+8^2}=\sqrt{16(2)+64}=\sqrt{32+64}=\sqrt{96}=\sqrt{16(6)}=4\sqrt{6}$

Finally, the surface area of the right circular cone is

$A_s=\dfrac{1}{2}cL+\pi r^2=\dfrac{1}{2}(2)(\pi)(4\sqrt{2})(4\sqrt{6})+\pi (4\sqrt{2})^2=32\pi \sqrt{3}+32\pi$