What is the least perimeter of a rectangle with an area of 32 square feet

Area of a rectangle=length x width.
length=x
width=y

xy=32   ⇒y=32/x

perimeter of a rectangle=2x+2y
P(x,y)=2x+2y
P(x)=2x+2(32/x)
P(x)=2x+64/x
P(x)=(2x²+64)/x

1) we calculate the first derivative
y=u/v   ;     y´=(u`v-v´u) / v²
P`(x)=(4x*x-1(2x²+64)] / x²
P´(x)=(4x²-2x²-64) / x²
P´(x)=(2x²-64) / x²

2)we equalize the first derivative to "0" and we find out the value of x:
(2x²-64) / x²=0
2x²-64=0*x²
2x²-64=0
2x²=64
x²=64/2
x²=32
x=⁺₋√32
We obtein two possible values: 
x₁=-√32; this value is not valid.
x₂=√32   ;  

3)we calculate the second derivative:
P``(x)=[4x * (x²)-2x(2x²-64)] / x⁴
P´´(x)=(4x³-4x³+128x) / x⁴
P´´(x)=128/x³

4) we find out if x=√32 is a maximum or a minimum
P´´(√32)=128/(∛32)=40.317...>0;
Therefore, like P´´(√32)>0 at x=√32 we have a minimum

5) we find out the value of y.

y=32/x
y=32 / √32=(32*√32)/32=√32

Therefore: x=√32, y=√32, the rectangle is a square.

Perimeter=2x+2y
Perimeter=2√32+2√32=4√32

the least perimeter of a rectangle with an area of 32 ft² is 4√32 ft.

The smallest perimiter is when the sides are the same legnth

example
if you built a rectangle with area 20
if you built it wwith sides 20 and 1, then the perimiter would be 42
if you build it with sides 5 and 4, the perimiter is 18

so trry to get perimiters as clos as possible

get them equal

aera of rectangle=legnth times width
set legnth=width so we can get least perimiter
area of rectangle=legnth times legnth
area=legnth^2
32=legnth^2
sqrt both sides
√32=legnth
4√2=legnth

so since legnth=width, this rectangle is now a square (this is allowed becasue defenition of rectangle is 4 right angles, and all squares are rectangles, but that doesn't means all rectangles ar squares)

perimiter of square=4legnth
perimiter of this square=4*4√2=16√2

answer is 16√2 feet