Answer
Depth of the gutter ($x$) = 5
Maximum cross-sectional area = $50$ sq in
Work Step by Step
*Refer back to page 340 for a complete reference on calculating maximum and minimum values of quadratic functions*
The depth of the gutter, according to the image on the exercise, is $x$ and the cross-sectional area is given by the formula $Area = (x)(20-2x) = 20x - 2x^{2}$. We can find the maximum area by re-writing the equation as a quadratic function $f(x) = ax^{2} + bx + c$:
$$f_{Area} = -2x^{2} + 20x + 0$$
Since we know that, for quadratic functions with $a<0$, the maximum value of $f(x)$ can be found when $x=\frac{-b}{2a}$, we can calculate the following:
$$x_{maximum} = \frac{-b}{2a} = \frac{-(20)}{2(-2)} = \frac{-20}{-4} = 5$$
And we can plug the value into the function for the cross-sectional area that we previously constructed:
$$f_{area} = -2(5)^{2} + 20(5) = -50 + 100 = 50$$
