How can I find the maximum volume of the prism if its length is 1m and the surface area is a constant $S$?
Take the prism's isosceles base to have dimensions $a$ as the longest side, $b$ as the equal sides, and $\theta$ as the angle between $a$ and $b$.
By length, I mean the distance from end to end of the prism.
This prism has both bases closed and the side with area $a$ open, as it will be used to catch falling water. It will look similar to this, but triangular:
I have tried: $$V = \frac{1}{2} ab \sin \theta$$ $$S = ab \sin \theta +2b$$ $$S = 2V + 2b$$ $$V = \frac{S-2b}{2}$$
I have tried the same approach with a rectangle, and at this point I took the derivative to find the maximum volume. I do not understand how to do that here, since it seems the maximum volume is where $b = 0$, which doesn't make sense. If I have made a mistake, I can't find it.
HINT: For the prism we get
$$V=A\cdot 1m$$ with $A=\frac{1}{2}b^2\sin(\gamma)$ For the Surface we get
$Const=2bm^2+\frac{1}{2}b^2\sin(\gamma)$ where $b$ the two equal sidelength of the triangle, $a$ denotes the third side and $\gamma$ is the angle between the thwo sides $b$ ($m$ denotes the length in meter)