What is the maximum height of a snowman (consisting from 3 snowballs) when I have $\frac{4}{3}\pi$ cubic meters of snow Let balls $1,2,3$ have radia $r_1, r_2, r_3$. Then 
$V_1=\frac{4}{3}\pi r_1^{3}$
,$V_2=\frac{4}{3}\pi r_2^{3}$
and $V_3=\frac{4}{3}\pi r_3^{3}$.
I know that $V_1 + V_2 + V_3 =\frac{4}{3}\pi$.
From this I get function $$f(V_1,V_3) = 2\left(\left(\frac{3V_1}{4\pi}\right)^{\frac{1}{3}} + \left(\frac{3V_3}{4\pi}\right)^{\frac{1}{3}} + \left(1 - \frac{3(V_1 + V_3)}{4\pi}\right)^\frac{1}{3}\right)$$
But I do not know how to derivate it since we have learnt only derivations of functions with $1$ variable; therefore, I cannot find maximum. Teacher gave us a tip that we should use Jensen inequality but I do not know how to apply it for this problem.
Thanks for your help.

Really, the problem is to maximize $h=2(r_1+r_2+r_3)$ given that $r_1^3+r_2^3+r_3^3=1$.  You can find critical points by expressing
$$h(r_1,r_2) = 2 [ r_1+r_2 + (1-r_1^3-r_2^3)^{1/3})]$$
Consider
$$f(x,y) = x+y+(1-x^3-y^3)^{1/3}$$
$$f_x = 1 -\frac{x^2}{(1-x^3-y^3)^{2/3}}$$
$$f_y = 1 -\frac{y^2}{(1-x^3-y^3)^{2/3}}$$
Setting each of these to zero reveals that $x=y$, and thus, $r_1=r_2=r_3=r$, such that
$3 r^3=1 \implies r=3^{-1/3}$.  The max height is thus $6 r = 2 \cdot 3^{2/3}$.
