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

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

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

6(2x + 5y)6 (2x) = 12x6 (5y) = 30y12x + 30y