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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 that maximize the enclosed area are both 12.5 yards for length and width, making the enclosure a square. The maximum area that can be enclosed is 156.25 square yards.

Step by step solution

01

Formula for Perimeter of a Rectangle

The total length of the fencing represents the perimeter of a rectangle which is given by \(2*(Length + Width)\). We have 50 yards of fencing so \(2*(Length + Width) = 50\). We can simplify this by dividing both sides with 2 to give \(Length + Width = 25\) yards.
02

Express Length in terms of Width

We can express the Length in terms of Width from the above equation, by solving for Length. This gives us \(Length = 25 - Width\) yards.
03

Area of a Rectangle in terms of Width

The area \(A\) of a rectangle is given by \(Length*Width\). Substituting the expression of Length from the previous step into this yields \(A = Width*(25 - Width)\) or equivalently \(A = 25*Width - Width^2\) square yards.
04

Find Maximum Area using Calculus

To find the maximum area, we find the derivative of the area with respect to Width, set it to 0 and solve for Width. This gives us \(A' = 25 - 2*Width = 0\), solving for Width provides \(Width = 12.5\) yards.
05

Determine Length and Maximum Area

Substitute the Width, 12.5 yards, back into perimeter equation to find the Length, which is also \(25 - 12.5 = 12.5\) yards. Substitute Width and Length into the area equation to find the maximum area which gives \(A = 12.5 * 12.5 = 156.25\) square yards.
06

Verification of Maximum

To verify that 156.25 square yards is indeed the maximum area, we perform a second derivative test. The second derivative of the area equation is \(A'' = -2\), which is negative, confirming that the area is maximized.

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