$r$ – radius of base
$h$ – height of cone
$l$ – slant height of cone
$V$ – Volume of cone
$A$ – surface area of cone
First of all I myself don't know what's the meaning of phrase "given surface area", since it is written given I would assume it to be constant. I need to maximize the volume:
$$
V = \frac{\pi r^2 h}{3},
$$
$$
\frac{dV}{dr} = \frac {\pi}{3} \left(2rh + r^2 \frac{dh}{dr}\right).
$$
To obtain critical points set $dV/dr = 0$:
$$
2rh + r^2 \frac{dh}{dr} = 0,
$$
$$
\frac{dh}{dr} = -2\frac{h}{r}.
$$
Now, $A = \pi r \left( r + \sqrt{h^2 + r^2} \right)$.
Using $dA/dr = 0$ and $dh/dr = -2h/r$ we would get
$$
h = \sqrt{8}r,
$$
I have got the answer but how can I check whether it's a minimum volume or maximum volume.
For example if I put $h = \sqrt{8}r$ in the equation of volume we would get
$V = (\sqrt{8}/3) \pi r^3$, so if I double differentiate it
$d^2V/dr^2 = 2\cdot \sqrt{8} \cdot \pi \cdot r^2 $, no matter what value or $r$ I put I always get a positive value which means $h = \sqrt{8}r$ represents a minimum volume.
Please spot out my mistake as I'm myself unable to do it.
Best Answer
Hint
For such a cone, we have
$$V=\frac 1 3 \pi r^2h \qquad \text{and} \qquad A=\pi r \left(r + \sqrt{r^2+h^2}\right)$$
Since $A$ is given, extract $h$ $$h=\frac{\sqrt{A} \sqrt{A-2 \pi r^2}}{\pi r}$$ Replace in $V$ and ... continue