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

Answer: 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