Problem 72 - College Algebra

Question

A rain gutter is made from sheets of aluminum that are 12 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?

Step-by-Step Solution

Answer

The depth of the rain gutter that results in the maximum cross-sectional area is 3 inches, and this maximum cross-sectional area is 18 square inches.

Step 1: Visualize and Understand the Problem

Visualize the rain gutter as a rectangular box with no top, which is formed by folding the two edges of an aluminum sheet upwards to create these sides. The width of the box (or gutter) is 12 inches minus twice its depth (because two equal parts of the sheet are folded upwards). The depth of the gutter corresponds to the height of the box, and we denote it by 'h'.

Step 2: Formulate an Equation for the Cross-Sectional Area

We can formulate an equation for the cross-sectional area (A) of the gutter as a function of the depth 'h'. Since the area of a rectangle is given by multiplying its length by its width, and because the length of the gutter is not specified, our shape is actually a rectangle with depth 'h' and its width is 12 inches - 2h (as each side of the rectangular box has width 'h'). So, the function A(h) = h(12 - 2h).

Step 3: Find the Maximum of the Cross-Sectional Area

To find the depth 'h' that maximizes A(h), we take the derivative of it with respect to 'h' and equate it to zero. This will give us the critical points where the maximum or minimum of A(h) occurs. The derivative A'(h) is given by the differentiation of A(h) = 12h - 2h², which is A'(h) = 12 - 4h. Setting A'(h) to 0 gives the equation 12 = 4h, or h = 3 inches.

Step 4: Confirm That This Is a Maximum

To confirm that this depth 'h' = 3 inches gives a maximum cross-sectional area and not a minimum, we take the second derivative of A(h), denoted by A''(h). This is given by differentiating A'(h) = 12 - 4h, which gives A''(h) = -4. Since this value is negative, it confirms that the depth h = 3 inches gives a maximum cross-sectional area.

Step 5: Compute the Maximum Cross-sectional Area

Finally, we can substitute h = 3 inches into our initial model for A(h) to find the maximum cross-sectional area. That is, A(h) = h(12 - 2h) = 3(12 - 2*3) = 3*6 = 18 square inches.