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Chapter 3: Problem 72

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?

Short Answer

Expert verified
The depth of the gutter that will maximize its cross-sectional area is 3 inches. And the maximum cross-sectional area of the gutter is 18 square inches.

Step by step solution

01

Model the situation using equations

Let's denote the depth (the height of the turned-up edge) as \(d\) and the width of the gutter (the distance between the turned-up edges) as \(w\). The total width of the aluminum sheet being used is 12 inches, so we have the equation \(2d + w = 12\). We can rearrange this to get \(w = 12 - 2d\). Let's proceed to calculate the cross-sectional area of the gutter, \(A\), which is a rectangle and can be found by the product of its depth and width. Therefore, \(A = d * w = d * (12 - 2d) = 12d - 2d^2 \).
02

Determine the maximum cross-sectional area

The maximum cross-sectional area can be found by getting the derivative of \(A\) where \(A'\) equals zero. The derivative of \(A\) is \(A' = 12 - 4d\). Setting this to zero we have \(12 - 4d = 0\) which gives \(d = 3\) inches. So, the depth of the gutter that gives a maximum cross-sectional area is 3 inches.
03

Verifying maximum value

We need to confirm if \(d = 3\) inches does indeed give us a maximum area. We could use the second derivative test. If the second derivative of the area, \(A''\), is less than zero at \(d = 3\), then \(d = 3\) is a maximum. The second derivative of \(A\) is \(A'' = -4\), which is indeed less than zero, so we have a maximum at \(d = 3\).
04

Calculating Maximum Cross-Sectional Area

To find the maximum area, we substitute \(d = 3\) into the formula for \(A\), \(A = 12d - 2d^2\). So, \(A_{max} = 12*3 - 2*3^2 = 36 - 18 = 18\). Therefore, the maximum cross-sectional area of the gutter is 18 square inches.

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