Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 68

Suppose that a rain gutter is to be constructed from a metal sheet of of width 30cm by bending up one-third of the sheet on each side through an angle $\theta$. How should $\theta$ be chosen so that gutter will carry the maximum amount of water?






The total Area of the gutter is

$
\begin{equation}
\begin{aligned}
A(\theta) &= \frac{1}{2} bh + 10h + \frac{1}{2} bh = 10h + bh\\
\\
A(\theta) &= 10(10 \sin \theta) + (10 \cos \theta) (10 \sin \theta)\\
\\
A(\theta) &= 100 \sin \theta + 100 \sin \theta \cos \theta
\end{aligned}
\end{equation}
$


Taking th derivative of $A$ by using Product Rule, we get...

$
\begin{equation}
\begin{aligned}
A'(\theta) &= 100 \cos \theta + 100 [ \sin \theta ( - \sin \theta ) + \cos \theta (\cos \theta)]\\
\\
A'(\theta) &= 100 \cos \theta + 100 \left[ - \sin ^2 \theta + \cos^2 \theta\right];
\end{aligned}
\end{equation}
$


Recall that $\sin^2 \theta + \cos^2 \theta = 1$

$
\begin{equation}
\begin{aligned}
A'(0) &= 100 \cos \theta + 100 [\cos ^2 \theta -1 + \cos^2 \theta]\\
\\
A'(0) &= 100 [ 2 \cos^2 \theta + \cos \theta - 1]
\end{aligned}
\end{equation}
$

When $A'(0) = \theta$,
$0 = 2 \cos^2 \theta + \cos \theta -1$

If we let $x = \cos \theta$,
$0 = 2x^2 + x -1$

By using Quadratic Formula,
$\displaystyle x = \frac{1}{2}$ and $x = -1$

So, we have
$\displaystyle \cos \theta = \frac{1}{2}$ and $\cos \theta = -1$

Thus,
$\displaystyle \theta = \frac{\pi}{3}$ and $\theta = \pi$
but $\theta < \pi$

Therefore, we can say that the gutter will carry out maximum water when $\displaystyle \theta = \frac{\pi}{3}$