Figure 19

A rain gutter is to be constructed using a piece of aluminum $12$ in wide. After marking a length of $4 \,{\rm in}$. from each edge, the piece of aluminum is bent up at an angle $\theta$, as illustrated in Figure 19. The area $A$ of a cross section of the opening, expressed as a function of $\theta ,$ is $$A(\theta) =16~\sin ~\theta ( \cos \theta +1)\qquad 0 \leq \theta \leq \dfrac{\pi }{2}$$

Find the angle $\theta$ that maximizes the area $A$. (This bend will allow the most water to flow through the gutter.)

Solution The function $A=A(\theta )$ is continuous on the closed interval $\left[ 0,\dfrac{\pi }{2}\right]$. To find the angle $\theta$ that maximizes $A$, we follow the three-step procedure.

Step 1 We locate all critical numbers in the open interval $\left( 0,\dfrac{\pi }{2}\right)$. $$\begin{eqnarray*} A^\prime (\theta ) &=& 16\sin \theta (-\!\sin \theta) +16\cos \theta ( \cos \theta +1) \hspace{12pc} \color{#0066A7}{{\hbox{Product Rule}}}\\[3pt] &=& 16[-\!\sin ^{2}\theta +\cos^{2}\theta +\cos \theta] \\[3pt] &=& 16[ ( \cos ^{2}\theta -1) +\cos ^{2}\theta +\cos \theta ] \hspace{15pc} \color{#0066A7}{{\hbox{$\kern-.5pt-\!\sin^2\theta=\cos^2\theta-1$}}}\\[3pt] &=& 16[2\cos^{2}\theta +\cos \theta -1] = 16(2\cos \theta -1) (\cos \theta +1) \end{eqnarray*}$$

The critical numbers satisfy the equation $A^\prime (\theta ) =0, 0<\theta<\frac{\pi}{2}$. $$\begin{eqnarray*} \begin{array}{rl@{ }l@{ }rl} 16(2\;\cos \theta -1) ( \cos \theta +1) &= 0 \\[3pt] 2\;\cos \theta -1 &= 0 & \quad\hbox{ or }\quad & \cos \theta +1 &=0 \\[3pt] \cos \theta &= \dfrac{1}{2} & \hbox{ }& \cos \theta &=-1\\[3pt] \theta &= \dfrac{\pi }{3} \;{\rm or}\; \dfrac{5\pi }{3} & \hbox{ }& \theta &= \pi \end{array} \end{eqnarray*}$$

Of these solutions, only $\dfrac{\pi }{3}$ is in the interval $\left( 0,\dfrac{\pi }{2}\right).$ So, $\dfrac{\pi }{3}$ is the only critical number.

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Step 2 We evaluate $A$ at the critical number $\dfrac{\pi }{3}$ and at the endpoints $0$ and $\dfrac{\pi }{2}$.

Step 3 If the aluminum is bent at an angle of $\dfrac{\pi }{3}$, the area of the opening is maximum. The maximum area is about $20.8 \,{\rm in}^{2}$.