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

P = 2L + 2W

The dimensions could be
1 by 32
2 by 16
4 by 8

The least perimeter is obtained when you have a square

let the side be s
s^2 = 32
s = √32 = 4√2

so the least perimeter is 8√2

by Calculus:
let the width be x
then the length is 32/x

P = 2x + 2y = 2x + 64/x
P' = 2 - 64/x^2 = 0 for a min of P

2 = 64/x^2
x^2 = 32
x = √32
etc (as above)

Oh ye

Oh yea

The dimensions 1 by 32,2 by 16, and 4 by 8 and if you calculate the perimeters here are the results:24,66 and 36. And as you can see the least is the rectangle with the dimension of 4 by 8 and the perimeter of 24.

oh yeahhhhh

The area of the shaded rectangle shown below is 15 ft.² what is the perimeter of the figure

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To find the least perimeter of a rectangle with an area of 32 square feet, we first need to determine the dimensions of the rectangle.

We know that the formula for the area of a rectangle is A = length × width.

In this case, the area (A) is given as 32 square feet.

To find the dimensions, we need to find two factors of 32 that will give us the minimum sum. Factors of 32 include: 1, 2, 4, 8, 16, and 32.

The two factors that have the smallest sum are 4 and 8. Let's assume 4 is the width and 8 is the length.

Now, we can find the perimeter of the rectangle by using the formula P = 2(length + width).

Substituting the values, we get P = 2(8 + 4) = 2 × 12 = 24 feet.

Therefore, the least perimeter of a rectangle with an area of 32 square feet is 24 feet.