One Problem at a Time

Application of Derivative: Optimization

Problem 229: If you have 60 meters of fencing and want to enclose a rectangular area up against a long, straight wall, what is the largest area you can enclose?

Solution: The perimeter of a rectangle is given by

(1)   \begin{equation*} P = 2L + 2W\end{equation*}

where L is the length of the rectangle and W is the width of the rectangle. In this case, we are given one side already so that we don’t considered. The perimeter then take the form

(2)   \begin{equation*} P = L + 2W \implies 60=2W + L \implies L = 60 - 2W. \end{equation*}

One can now find the area of the rectangle. That is,

(3)   \begin{align*} A & = L \cdot W \\ & = (60-2W) \cdot W \\ & = 60W -2W^2 .\end{align*}

We now consider A' = 0 to find the critical point.

(4)   \begin{align*} A & = 60 W - W^2 \\ A' & = 60 - 4W \\ 0 & = 60 - 4W \\ 4W & = 60 \\ W & = 15. \end{align*}

Using (2),

(5)   \begin{equation*} L = 60 - 2W = 60 - 2(15) = 60-30 = 30. \end{equation*}

Notice that

(6)   \begin{equation*} A' = 60-4W \implies A'' -4 < 0. \end{equation*}

By the second derivative test, we have an absolute maximum. Therefore, the largest area that can be enclosed is

(7)   \begin{equation*}A = L \cdot W = (15)(30) = 450 \ \text{m}^2. \end{equation*}

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