Maximize the volume of a block inscribed in hemisphere I want to find the maximum volume block with dimensions $a$, $b$ and $c$ incribed inside the hemisphere with radius $R$. I need to use lagrange multipliers. So far I found out that:
$R^2 = c^2 + \frac{a^2}{4} + \frac{b^2}{4}$ 
$V_{block} = abc$
$V_{hemisphere} = \frac{4}{6}\pi R^3 = \frac{4}{6}\pi(c^2 + \frac{a^2}{4} + \frac{b^2}{4})^{3/2}$ 
The equation for $R$ is taken from this picture
From here
$L = abc + \lambda\frac{4}{6}\pi(c^2 + \frac{a^2}{4} + \frac{b^2}{4})^{3/2}$
And now I have folowing equations:
$L_a = bc + \lambda\pi \frac{a}{2}(c^2 + \frac{a^2}{2} + \frac{b^2}{2})^{1/2} = 0$
$L_b = ac + \lambda\pi \frac{b}{2}(c^2 + \frac{a^2}{2} + \frac{b^2}{2})^{1/2} = 0$
$L_c = ac + \lambda\pi 2c(c^2 + \frac{a^2}{2} + \frac{b^2}{2})^{1/2} = 0$
$\frac{4}{6}\pi(c^2 + \frac{a^2}{4} + \frac{b^2}{4})^{3/2} = 0$ 
Which seems unsolveable for me. Did I make mistake somewhere?
Thx for help.

The formula for $V_{\textrm{hemisphere}}$ is unnecessary---this information is already encoded in the radius $R$ that appears in the constraint. In any case you're aiming to maximize the quantity
$$V = a b c$$
subject to the condition $$R^2 = \frac{a^2}{4} + \frac{b^2}{4} + c^2 .$$
Note that by parameterizing our blocks in a slightly different way, namely with $a = 2 A$, $b = 2 B$, $c = C$, we get the similar problem of trying to maximize the quantity
$$V = 4 A B C$$
subject to the condition $$R^2 = A^2 + B^2 + C^2 .$$
This formulation has the advantage of being symmetric in $A, B, C$.
