Q. No. 7 - Precalculus Enhanced with Graphing Utilities

Question

Enclosing a Rectangular Field: David has 400 yards of fencing and wishes to enclose a rectangular area.

(a) Express the area A of the rectangle as a function of the width w of the rectangle.

(b) For what value of w is the area largest?

(c) What is the maximum area?

Step-by-Step Solution

Answer

(a) The area A of the rectangle as a function of the width w of the rectangle is $A=-w^2+200w$.

(b) The area is the largest when $w=100$.

(c) The maximum area is 10,000 square yards.

Part (a) Step 1. Given information.

Given that  David has 400 yards of fencing and wishes to enclose a rectangular area.

Part (a) Step 2. The area as a function of width of the rectangle.

The perimeter of the rectangular area is $400$ yards.

Let the width of the rectangle be w which means

$$\begin{gathered} 2(l+w)=400 \\ l+w=200 \\ l=200-w \end{gathered}$$

The area A as a function of the width of the rectangle is

$$\begin{gathered} A=lw \\ A=(200-w)w \\ A=200w-w^2 \\ A=-w^2+200w \end{gathered}$$

Part (b) Step 1. The value of w for the largest area.

The function A is a quadratic function with $a=-1,b=200$, and $c=0$. Because $a<0$, the vertex is the highest point on the parabola.

The area A is a maximum when the width w is

$$\begin{gathered} w=-\frac{b}{2a} \\ =-\frac{200}{2(-1)} \\ =100 \end{gathered}$$

Part (c) Step 1. Substitute x = 100 in A = - w 2 + 200 w .

We get

$$\begin{gathered} A=-{\left(100\right)}^2+200\left(100\right) \\ =-10000+20000 \\ =10000 \end{gathered}$$

The maximum area is 10,000 square yards