Volume of cylinder. Inscribe a cylinder with largest possible volume into a cone whose height is $h$ and radius of base is $R$, so that the lower base of cylinder lays on base of cone, and the upper base touches cone from the inside. Find the volume of such cylinder using extreme values of multivariable functions.
This seems more like a problem for double or triple integrals, but i have no idea how to solve it using extreme values of multivariable functions, any ideas?
 A: Let $r $ be the radius of the cylinder. Then its height is, by triangle similarity, $(R-r )h/R $. So the cylinder volume is 
$$ \frac {\pi h}{R} r^2 (R-r )$$
To maximize it, we can take the derivative and set it to zero, getting
$$ \frac {\pi h}{R} r (2R-3r )=0$$
The derivative is zero for $r=0$ and $r=2/3 R $, is positive between these two values, and is negative for $r >2/3 R $. From this we get  that the maximum volume is obtained for $$r=2/3 \, R $$
This gives a cylinder height of $h/3$, so the resulting volume is
$$  4/27\, \pi  R^2h   $$
EDIT: to solve this problem using the estreme values of multivariate functions, simply call again $r $ the radius of the cylinder and $k $ its height. Consider the function $\pi r^2 k $, expressing the cylinder volume, as a two-variable function defined in the region bounded by $0 \leq r \leq R $ and $ 0 \leq k \leq (R-r)h/R$. Equalizing the two derivatives (in $r $ and in $k $) to zero and solving the system we get the critical value $x=0$, which is a local minimum. Considering then the boundaries, and in particular that given by $ k =(R-r)h/R$, you are left with the same equation reported above, whose derivative in $r $ is zero for $r=2/3 \, R $, corresponding to the local maximum.
