Answer
$\theta=\frac{\pi}{6}$
Work Step by Step
Step 1. The volume of the trough can be expressed as $V=20A$ where $A$ is the area of the side plate.
Step 2. Using the figure given in the Exercise, we have $A=1\times cos\theta+2\times\frac{1}{2}sin\theta cos\theta\ $ where the first term is the area of the middle rectangle and the second term is the area of the two side triangles.
Step 3. With $V=20(cos\theta+sin\theta cos\theta)$, we have $V'=20(-sin\theta+cos^2\theta-sin^2\theta)=20(1-sin\theta-2sin^2\theta)$
Step 4. Letting $V'=0$, we have $sin\theta=-1,\frac{1}{2}$ with $0\lt\theta\lt \frac{\pi}{2}$, thus $\theta=\frac{\pi}{6}$ and $V=20(cos\frac{\pi}{6}+sin\frac{\pi}{6} cos\frac{\pi}{6})=15\sqrt 3ft^3$
Step 5. Check $V''=20(1-cos\theta-2sin\theta cos\theta)|_{\theta=\frac{\pi}{6}}=20(1-\sqrt 3)\lt0$
This indicates that the function is concave down with a maximum at $\theta=\frac{\pi}{6}$.
