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.

Your objective functio is $V=abc$ and the constrain is $ R^2=\frac{a^2}{4} + \frac{b^2}{4} + c^2$ then your expression should be
$$L=abc+\lambda\left(\frac{a^2}{4} + \frac{b^2}{4} + c^2 -R^2\right)$$
and then
$$L_a=bc-\lambda\frac{a}{2}=0\to \lambda a=2bc$$
$$L_b=ac-\lambda\frac{b}{2}=0\to \lambda b=2ac$$
$$L_c=ab-2\lambda c=0\to \lambda c=\frac{ab}{2}$$
Can you finish?
