Maxima and minima problem A conical tent of the given capacity (volume) has to be constructed. Find the ratio of the
height to the radius of the base so as to minimise the canvas requried for the tent.
 A: Hint
The lateral surface of the cone is given by $A = \pi  r \sqrt{h^2+r^2}$ and the volume is given by $V=\frac{1}{3} \pi  h r^2$.
Since you look for this ratio, let us call $x=\frac{h}{r}$; so, we can rewrite $A=\pi  r^2 \sqrt{x^2+1}$ and $V=\frac{1}{3} \pi  r^3 x$.
Since $V$ is given,we can extract $r$ from $V$; this leads to $r=\sqrt[3]{\frac{3 V}{\pi x}}$. Then, back to $A$, we have, after a few simplifications, $$A= \sqrt{3 \pi V}  \sqrt{\frac{x^2+1}{x}}$$ So, finding the minimum of $A$ means finding the minimum of $$y=\sqrt{\frac{x^2+1}{x}}$$ This minimum will correspond to the value of $x$ which cancels the derivative of $y$ with respect to $x$.
I am sure you can easily take from here.
A: the required ratio of height and radius of the cone will be 
h/r=2^0.5
A: Write down expressions for the surface area and volume of the cone in terms of the radius of the base and the height. 
The problem gives you $volume=c$ for some given constant $c$. Solve this equation for height and substitute into the equation for surface area. This gives you surface area as a function of radius.
Now, you can take the derivative, set it equal to $0$ and apply the first/second derivative test to find a minimum. 
A: V = 1/3 pi r^2 h
SA = pi r s + pi r^2
Since volume is fixed and is a function of the radius of the base and the height of the cone, it suffices to find just one of those values (it will give us the other and therefore the ratio). Since those are the two variables we are concerned with, start by getting rid of that nasty slant height, s, in the second equation. Use the Pythagorean theorem to write it as root(r^2 + h^2).
I'll solve the first equation for h:
h = (3V)/(pi r^2)
plug that into the second equation and we have a (nasty) function for surface area in terms of r. 
SA = pi r root[r^2 + (3V/pi r^2)^2] + pi r^2
There's your f(r), now just differentiate it, set it to zero, solve for r values and check them to see which yields the minimum surface area.
