SOLUTION: Sam has 3000 feet of fencting 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)

Algebra ->  Length-and-distance -> SOLUTION: Sam has 3000 feet of fencting 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)      Log On


   

Question 153414: Sam has 3000 feet of fencting 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?
Answer by orca(409) About Me  (Show Source):
You can put this solution on YOUR website!
First we need to express the width of the rectangle in terms of x.
The perimeter of the rectangle is:
Perimeter = 2*length + 2*width.
So width = (perimeter - 2*length)/2
Substituting length = x and perimeter = 3000, we have:
width+=+%283000-2x%29%2F2=1500-x
So the area of the rectangle is
A=length%2Awidth
=x%2A%281500-x%29
=1500x-x%5E2%29
To find the maximum value of the quadratic expression, we can use either the formula or the completing square method. Here the second method is used.
A+=+1500x-x%5E2
=-x%5E2%2B1500x
=-%28x%5E2-1500x%29
=-%28x%5E2-2%2A750x%29
=-%28x%5E2-2%2A750x%2B750%5E2-750%5E2%29
=-%28%28x%5E2-2%2A750x%2B750%5E2%29-750%5E2%29
=-%28%28x-750%29%5E2-750%5E2%29
=-%28x-750%29%5E2%2B750%5E2
=750%5E2-%28x-750%29%5E2
750%5E2-%28x-750%29%5E2 has maximum value 750%5E2 when %28x-750%29%5E2=0.
For %28x-750%29%5E2=0, x must be equal to 750.
So when x = 750, A has the maximum value 750%5E2.