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

answer: the answer is (b) amie should have multiplied

step-by-step explanation: given a sphere and a cylinder with same radius and height.

let, 'r' be the radius of the cylinder and sphere and 'h' be the height of the cylinder. according to the question,

r = h.

now, the volume of the sphere will be

and the volume of the cylinder will be

therefore, we have

thus, ammie should have multiplied the volume of the sphere

126 maybe. sorry