Answer
Depth is $\text{5 inches}$ and the maximum cross-sectional area is $\text{50 square inches}$.
Work Step by Step
Let us assume the depth of the gutter be x.
Then, the length is given by $20-2x$.
And, the cross-sectional area of the gutter is
$\begin{align}
& A=\left( \text{depth }\!\!\times\!\!\text{ length} \right) \\
& =x\left( 20-2x \right)
\end{align}$.
So,
$\begin{align}
& A=x\left( 20-2x \right) \\
& =20x-2{{x}^{2}} \\
& =-2{{x}^{2}}+20x
\end{align}$
Which is the equation of a parabola opening downwards. Compare it with the standard equation, $a{{x}^{2}}+bx+c$ , get
$a=-2$ and $b=20$.
It is known that for the standard equation, $a{{x}^{2}}+bx+c$ , if $a<0$ then the maximum value occurs at $x=-\frac{b}{2a}$.
Thus, the maximum of A, that is maximum value of cross-sectional area, will occur at
$\begin{align}
& x=-\frac{b}{2a} \\
& =-\frac{20}{2\left( -2 \right)} \\
& =-\frac{20}{\left( -4 \right)} \\
& =5
\end{align}$
Therefore, the depth is $x=5$ inches.
$\begin{align}
& \text{Length}=20-2\left( 5 \right) \\
& =20-10 \\
& =10
\end{align}$
And, the maximum cross-sectional area is
$\begin{align}
& {{A}_{\max }}=x\left( 20-2x \right) \\
& =5\left( 20-2\left( 5 \right) \right) \\
& =5\left( 20-10 \right) \\
& =50
\end{align}$
Hence, when the depth of the gutter is $\text{5 in}$ the maximum cross-sectional area that can be enclosed is $\text{50 square inches}$.
