a rain gutter is made from sheets of aluminum that are 20 inches wide by turning up the edges to form right…

Explanation:

Step1: Set up the area function

Let the depth of the gutter be $x$ inches. The base of the rectangular - cross - section of the gutter is $20 - 2x$ inches and the height (depth) is $x$ inches. The cross - sectional area $A(x)$ of a rectangle is given by the formula $A(x)=x(20 - 2x)=20x-2x^{2}$.

Step2: Find the derivative of the area function

Using the power rule, if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. For $A(x)=20x-2x^{2}$, $A^\prime(x)=\frac{d}{dx}(20x-2x^{2})=20 - 4x$.

Step3: Find the critical points

Set $A^\prime(x) = 0$ to find the critical points. So, $20 - 4x=0$. Solving for $x$: [ \begin{align*} 20-4x&=0\ 4x&=20\ x& = 5 \end{align*} ]

Step4: Determine if it's a maximum

Find the second - derivative of $A(x)$. $A^{\prime\prime}(x)=\frac{d}{dx}(20 - 4x)=-4$. Since $A^{\prime\prime}(5)=-4\lt0$, by the second - derivative test, $x = 5$ is a point of maximum.

Step5: Find the maximum area

Substitute $x = 5$ into the area function $A(x)$. $A(5)=5\times(20 - 2\times5)=5\times(20 - 10)=50$ square inches.

Answer:

The cross - sectional area is maximized when the depth of the gutter is $5$ inches. The maximum cross - sectional area is $50$ square inches.