Question

A rain gutter is made from sheets of aluminum that are 22 inches wide by turning up the edges to form right angles. Determine the depth of the gutter that will maximize its​ cross- sectional area and allow the greatest amount of water to flow. What is the maximum​ cross-sectional area?

Answer 1

So hmm notice the picture below

thus, the perimeter of the gutter is just w + w + l, or 2w +l

thus $\bf 22=2w+l\implies 22-2w=l \\\\\\ \textit{area of a rectangle}\\\\ A=lw\qquad 22-2w=l\implies A(w)=(22-2w)w \\\\\\ A(w)=22w-2w^2$

take the derivative of A(w), zero it out, check any critical points in the interval of (0, 22), for a maxima

recall, the aluminum sheet is just 22inches long, thus, whatever the "depth" or "w" is, has to be more than 0, and less than 22