Hello leprofece,
Let's first draw a diagram:
View attachment 1655
The area $A$ of the trapezoid is the area of the middle rectangle and the areas of the right triangles on either side:
$$A=hs+2\left(\frac{1}{2}hs\cos(\theta) \right)=hs\left(1+\cos(\theta) \right)$$
Now we have $$h=s\sin(\theta)$$ hence:
$$A(\theta)=s^2\sin(\theta)\left(1+\cos(\theta) \right)$$
Differentiating with respect to $\theta$ and equating the result to zero, we find:
$$A'(\theta)=s^2\left(-\sin^2(\theta)+\cos(\theta)\left(1+\cos(\theta) \right) \right)=s^2\left(2\cos^2(\theta)+\cos(\theta)-1 \right)=s^2\left(2\cos(\theta)-1 \right)\left(\cos(\theta)+1 \right)=0$$
Since $0<s$, and $0\le\theta<\pi$ this implies:
$$\cos(\theta)=\frac{1}{2}\,\therefore\,\theta=\frac{\pi}{3}$$
Using the first derivative test, we find:
$$A'(0)=2s^2>0$$
$$A'\left(\frac{\pi}{2} \right)=-s^2<0$$
Thus we conclude the critical value $$\theta=\frac{\pi}{3}$$ is at a maximum for the area, and we can easily see this gives us a trapezoid that is a semi-hexagon.
