Chapter 3.4, Problem 8AYU

Enclosing a Rectangular Field Beth has 3000 feet of fencing available to enclose a rectangular field.

(a) Express the area $A$ of the rectangle as a function of $x$. where $x$ is the length of the rectangle.

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

(c) What is the maximum area?

To determine

To calculate:

Express the area of the rectangle as a function of width w.

For what value of w is the area largest.

What is the maximum area?

Answer to Problem 8AYU

Solution:

$A=1500x-x^2$

750 yards

5,62,500 square yards.

Explanation of Solution

Given:

Beth has 3000 yards of fencing and wishes to enclose a rectangular area.

Formula used:

Perimeter of a rectangle is twice the sum of its length and width.

Area of a rectangle is $A=lw$, where l is the length and w is the width.

Calculation:

Let the length of the rectangle be x and width be w.

The 3000 yards of fencing can be considered as the perimeter of the rectangle.

Therefore, we get

$2(x+w)=3000$

$\Rightarrow x+w=1500$

$\Rightarrow w=1500-x$

Now, we have to find the area of the rectangle.

Thus, we get the area as

$A=(1500-x)x=1500x-x^2$

a. The area of the rectangular fence is A=1500x-x^2.

b. The equation of area is a quadratic equation with $a=-1,b=1500,c=0$. Since a is negative, the vertex is the maximum point of the equation.

Thus, we get the maximum point at $-\frac{b}{2a}=-\frac{1500}{2(-1)}=750$

Therefore, the area is maximum when the length is 750 yards.

c. The maximum area of the rectangle is

$A\left(-\frac{b}{2a}\right)=A(750)=1500(750)-(750{)}^2$

$=562500$

The maximum area of the rectangle is 5,62,500 square yards.