You have 64 feet of fencing. If you are wanting to make a rectangular pen, what is the greatest area you can enclose?

Answer 1

A=lw

P=2l+2w

64=2l+2w

divide both sides by 2

32=l+w

solve for one or the other variable

32-l=w

A=lw

A=l(32-l)

A=32l-l^2

Since in a geometry course you don't study derivatives, you can do the following:

-b/2a is the maximizing point

-32/2(-1)=-32/(-2)=16

A=32(16)-16^2

A=512-256

A=256 square feet which is the maximum area

You want a square that is 16' x 16'

You can check to see if this is the maximum area.

Try a rectangle that is 15' by 17'

A=15*17

A=255 square feet which is less.

Try a rectangle that is 14' by 18'

A=14*18

A=252 square feet which again is less

If you continue doing this you will get smaller and smaller areas.

This method eliminates the need for derivatives.

Answer 2

64/4 = 16 feet per side of a square

max area = 16^2 = 256 square feet

a square has maximum area for a fixed perimeter of a quadrilateral or rectangle

or you could solve the problem using calculus, derivatives, optimization

64 feet = the perimeter = 2(L+W)

64 = 2(L+W)

32 = L+W

W = 32-L

Area = A = LW = L(32-L) = 32L -L^2

A= 32L -L^2

take the derivative with respect to L, set = 0 and solve for L

A' = 32-2L = 0

2L = 32

L= 16

W = 32-L = 32-16 =16

Max A = LW = 16(16) = 256 ft^2