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Chapter 3: Problem 67

You have 50 yards of fencing to enclose a rectangular region. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?

Short Answer

Expert verified
The dimensions of the rectangle that maximize the enclosed area are length = 12.5 yards and width = 12.5 yards, and the maximum area is 156.25 square yards.

Step by step solution

01

Formulate the Perimeter Constraint

Since the perimeter of the rectangle consists of two lengths and two widths, and you know that the total perimeter is 50 yards, therefore, it can be formulated as \(2L + 2W = 50\) or \(L + W = 25\). This formula represents the perimeter constraint.
02

Express Width in terms of Length

By rearranging the equation from step 1, we get \(W = 25 - L\). This equation expresses the width in terms of the length.
03

Apply Rectangle Area Formula

The area \(A\) of a rectangle is calculated as \(A = L \times W\). If we replace \(W\) from step 2, we get \(A = L \times (25 - L)\) or \(A = 25L - L^2\). This formula represents the area of the rectangle in terms of the length.
04

Differentiate and find critical points

To maximize the area, differentiate the equation of \(A\) with respect to \(L\) and set the result equal to zero: \(A' = 25 - 2L = 0\), therefore \(L = 12.5\). This gives the length of the rectangle that generates the maximum area. Remember that critical points can also be found where derivative is not defined or at the boundaries of the domain, but here, as \(L\) is always defined and there is no specific given domain, just checking \(L = 12.5\) suffices.
05

Validate Maxima

To confirm whether \(L = 12.5\) yields a maximum, take second derivative and substitute \(L = 12.5\). The second derivative is \(A'' = -2\) which is less than 0, hence, confirming that \(L = 12.5\) due to negative concavity ensures maximum area.
06

Find Width

Substitute \(L = 12.5\) into the equation of \(W\) from step 2, we get \(W = 25 - 12.5 = 12.5\). So, the width is also 12.5 yards.
07

Find Maximum Area

Substitute \(L = 12.5\) into the equation of \(A\) from step 3, we get \(A max = 12.5 \times (25 - 12.5) = 156.25\). So, the maximum area is 156.25 square yards.

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