Question

Find the dimensions of the rectangular corral producing the greatest enclosed area given 320 feet of fencing. (Assume that the length is greater than or equal to the width.)

Answer 1

Answer:

The dimensions of the rectangular corral producing the greatest enclosed area  is a square of 80 ft x 80 ft

Step-by-step explanation:

Let

x -----> the length of the rectangular corral in feet

y -----> the width of the rectangular corral in feet

we know that

The area of the rectangular corral is equal to

$A=xy$ -----> equation A

The perimeter of the rectangular corral is equal to

$P=2(x+y)$

$P=320\ ft$

so

$320=2(x+y)$

Simplify

$160=(x+y)$

$y=160-x$ -----> equation B

substitute equation B in equation A

$A=x(160-x)$

$A=-x^2+160x$

This is a vertical parabola open downward

The vertex is a maximum

The x-coordinate of the vertex represent the length for the maximum area

The y-coordinate of the vertex represent the maximum area

Convert the quadratic equation in vertex form

$A=-x^2+160x$

Factor -1

$A=-(x^2-160x)$

Complete the square

$A=-(x^2-160x+80^2)+80^2$

$A=-(x^2-160x+80^2)+6,400$

Rewrite as perfect squares

$A=-(x-80)^2+6,400$

The vertex is the point (80,6,400)

so

$x=80\ ft$

The maximum area is 6,400 ft^2

Find the value of y

$y=160-x$  ----> $y=160-80=80\ ft$

therefore

The dimensions of the rectangular corral producing the greatest enclosed area  is a square of 80 ft x 80 ft