Find the dimensions of the rectangle with perimeter 64 inches that has maximum area, and then find the maximum area.

  the maximum area is 256 in^2

Step-by-step explanation:

It is well-known that the rectangle of given perimeter with maximum area is a square. A square with a perimeter of 64 inches has a side length of 64/4 = 16 inches. The area of that square is the square of the side length, ...

   (16 in)^2 = 256 in^2

Derivation of the well-known fact

Let P represent the perimeter of a rectangle with side lengths x and (P/2 -x). The area is the product of these dimensions:

  A = x(P/2 -x)

This is the equation of a downward-opening parabola with zeros at x=0 and at x=P/2. The vertex (maximum) is halfway between these zeros, at x=P/4. Hence, the dimensions of the rectangle are ...

  x = P/4 . . . . . . . . . . . length

  (P/2 -P/4) = P/4 . . . . width

The length and width are the same, and both are 1/4 of the perimeter, so the rectangle is a square. Its area in terms of perimeter is ...

   A = (P/4)^2 = P^2/16

40 minutes

step-by-step explanation:

add the time and multiply by 2

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!