Enclosing Rectangular Field

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 3, Section 3.4

See attachment

[Eigenvalue]

A=lw
P=2(l+w)=400
l+w=200

1. Write w in terms of l:
l=200-w

2. Substitute l into the area equation
A=w(200-w)=-w²+200w

3. To find the value for which the value of w is the largest, find the x-coordinate of the vertex
w=[tex]\frac{-b}{2a}[/tex]=[tex]\frac{-200}{2(-1)}[/tex]=100

4. Input the value of w into the area equation
A=(-100)²+200(100)=-10000+20000=10000

[Eigenvalue]

Note: for questions that involve finding the maximum area of a rectangle, given the perimeter n, then the maximum area is a square with side lengths

A=lw
n=2(l+w)
l+w=[tex]\frac{n}{2}[/tex]

1. Rewrite l in terms of w
l=[tex]\frac{n}{2}[/tex]-w

2. Substitute l into the area formula
A=w([tex]\frac{n}{2}[/tex]-w)=-w²+[tex]\frac{n}{2}[/tex]w

3. Find the x-coordinate of the vertex to find the width for the maximum area
w=[tex]\frac{-b}{2a}[/tex]=[tex]\frac{-n/2}{2(-1)}[/tex]=[tex]\frac{n}{4}[/tex]

4. Find the length
l+w=[tex]\frac{n}{2}[/tex]
l=[tex]\frac{n}{4}[/tex]
Sidenote: as the length and width are equal, the rectangle is a square

5. Find the area by substituting in the value for w
A=([tex]\frac{n}{4}[/tex])([tex]\frac{n}{2}[/tex]-[tex]\frac{n}{4}[/tex])

=([tex]\frac{n}{4}[/tex])([tex]\frac{n}{4}[/tex])

=[tex]\frac{n²}{16}[/tex]