Find the dimensions of a rectangle with perimeter 64 m whose area is as large as possible.

The dimensions would be 16 because 16+16+16+16=64 so if you wanted to fin the are it would be 16*16=256 squared inches

I have found that when the length=width for any rectangle, the area is maximized

note: all squares are rectangles but not all rectangles are squares, so don't tell me that a rectangle can't be a square because rectangle is defined as having 4 right angles and a square is a special rectangle with 4 right angles and 4 equilateral sides

so

64=p=2 (L+W)

64=2 (L+W)

we assume L=W for max area

64=2 (L+L)

64=2 (2L)

64=4L

divide both sides by 4

16=L

area=LW or L^2 for a square

area=16*16=256 m²

if you wanted calculus way

64=P=2 (L+W)

divide by 2

32=L+W

32-L=W

area=LW

subsitute 32-L for W

area=L (32-L)

area=32L-L²

take derivitive of both sides

area dy/dL=32-2L

find where it equals 0

0=32-2L

2L=32

L=16

is it a max or min?

dy/dL is negative when L=17 and positive when L=15

so it goes up then down

it is a max

L=16

32-L=W

32-16=W

16=W

area=16*16=256

area=256 m²