# Peter has 800 yards of fencing to enclose a rectangular area. Find the dimensions of the...

## Question:

Peter has 800 yards of fencing to enclose a rectangular area. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?

## Area of a Rectangle; Maxima and Minima:

{eq}\\ {/eq}

Here we have given a rectangular shape in which the length and the width are related to each other through an algebraic expression (Perimeter of the rectangle) and we have to maximize the area of the rectangle. First of all, we will form the function of the area in terms of its length then using the concept of maxima and minima, we will determine such value of length that will correspond to the maximum value of the area.

$$\text {Critical Points} \; \; \; \Longrightarrow \; \; \biggr( \dfrac {df(x)}{dx} \biggr) = 0 $$

$$\text {Correspond to Maxima} \; \; \; \Longrightarrow \; \; \; \biggr( \dfrac {d^{2}f(x)}{dx^{2}} \biggr) < 0 $$

## Answer and Explanation:

{eq}\\ {/eq}

Let us assume:

The length of the rectangular area is {eq}\; = \text {L} \; {/eq} yards.

The width of the rectangle is {eq}\; = \text {W} \; {/eq} yards.

We have provided the information that the total length of the fencing is 800 yards.

{eq}2 \biggr( \text {L} + \text {W} \biggr) = 800 \\ \text {L} + \text {W} = 400 \\ \text {W} = 400 - \text {L} \; \; \; \cdots \cdots \; \; \; (1) {/eq}

{eq}\text {Area} = \text {L} \times \text {W} \; \; \; \cdots \cdots \; \; \; (2) {/eq}

Now using equation (1) and (2), we get the following:

{eq}\text {A} = \biggr( 400 - \text {L} \biggr) \times \text {L} \\ \text {A} = 400 \text {L} - \text {L}^{2} \\ {/eq}

Now determine the critical points of the above function using the first derivative:

{eq}\dfrac {d\text {A}}{d \text {L}} = 0 \\ \dfrac {d \text {A}}{d \text {L}} = \dfrac {d \biggr( 400 \text {L} - \text {L}^{2} \biggr)}{d \text {L}} \\ \dfrac {d\text {A}}{d\text {L}} = 400 - 2 \text {L} = 0 \\ \text {L} = \biggr( \dfrac {400}{2} \biggr) = 200 \; \text {yards} {/eq}

Now using the second derivative test, we will check that the value of {eq}\text {L} = 200 {/eq} will correspond to either maxima or minima of the function.

{eq}\dfrac {d^{2}\text {A}}{d \text {L}^{2}} = \dfrac {d \biggr( 400 - 2 \text {L} \biggr)}{d \text {L}} \\ \dfrac {d^{2} \text {A}}{d \text {L}^{2}} = - 2 < 0 {/eq}

Hence the value of {eq}\text {L} = 200 \; {/eq} yards correspond to the maximum value of the function.

Now put the value of {eq}\text {L} = 200 {/eq} yards in equation (1) in order to get the value of {eq}\text {W} {/eq}:

{eq}\text {W} + 200 = 400 \\ \text {W} = 200 \; \text {yards} {/eq}

The dimensions of the rectangle that maximize the enclosed area are given below:

{eq}\Longrightarrow \boxed {(\text {L}, \; \text {W}) = (200, \; 200) \; \text {yards}} {/eq}

The maximum enclosed area is {eq}\; = 200 \times 200 = 40000 \; \text {yards}^{2} {/eq}

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from CAHSEE Math Exam: Tutoring Solution

Chapter 10 / Lesson 12