Calculate the optimal angle for the roof of this dog house. I'm new to optimization and I'm stuck with the following problem:

Given four square planks 10 cm long each one, without cutting or changing any shape, we use two of them to make the walls and the other two to build the roof (by forming an isosceles triangle). What should be the optimal size of the angle $\alpha$ between the walls and the roof for the house to have the biggest possible volume?

I've spent to much time with this, even though I'm pretty sure the answer is quite simple.

After many trials, the only thing I imagined was that the total volume of the house could be expressed as $$V=2a^2b+abh=200b+10b(\sqrt{100-b^2})$$
whose derivative find a positive local maximum at $b = 5\sqrt{2}\sqrt[4]{3}$, leading to an optimal angle around $167º$, which is definitely absurd.

Thank you in advance.

EDIT

I'll explain my $a$, $b$ and $h$. If we divide in two de base of the prism that forms the roof, we get two right triangles. $a$ is the the side of each plank, $b$ is the base of both right triangles and $h$ is the hight of those.
We can treat this by maximizing the cross sectional area A.

$A = (100 - x^2)^{.5} (20 + x)$ where $x$ is the appex height of the roof

$dA/dx = \frac{-x(20+x)}{(100-x^2)^{.5}} + (100-x^2)^{.5}$

A is max when $dA/dx = 0$

$0 = \frac{-x(20+x)}{(100-x^2)^{.5}} + (100-x^2)^{.5}$

$\frac{x(20+x)}{(100-x^2)^{.5}} = (100-x^2)^{.5}$

$20x + x^2 = 100 - x^2$

$2x^2 + 20x - 100 = 0$

$x^2 +10x - 50 = 0$

This yields a positive $x = 3.660254$

Angle $\alpha = 90 + \sin^{-1}(3.660254/10)$

$\alpha = 111.47$ deg
My algebra gives b=√(√7500) or 9.30604 cm, (same as your answer) leading to an (internal) angle of 111.47°--which agrees with @Bence's result.