On a rectangular piece of cardboard with perimeter 10 inches, three parallel and equally spaced creases are made (see figure 1). the cardboard is then folded along the creases to make a rectangular box with open ends (see figure 2). letting x represent the distance (in inches) between the creases, use the aleks graphing calculator to find the value of x that maximizes the volume enclosed by this box. then give the maximum volume. round your responses to two decimal places.

The volume of a shape is the amount of space in it.

The given parameter is:

Let the height of the cardboard be h.

So, the perimeter of the cardboard (see attachment) is:

Substitute

Divide through by 2

Make h the subject

The volume of the cardboard is then calculated as:

Substitute

Expand

Differentiate

Set to 0

Rewrite as:

Divide through by -x

Divide through by 12

Recall that:

So, we have:

Take LCM

Hence,

Read more about volumes at:

link

The width of the cardboard (in inches) will be 4x, and its length will be (5-4x) so the perimeter adds up to 10 in. Then the volume is the product of the area of the open ends (x²) and the length (5-4x).

The graphing calculator shows that volume is maximized when the value of x is 0.833, or 5/6 inch. The corresponding length is 5-4x = 5/3 inch, so the volume is ...

... v = (5/6)²×(5/3) = 125/108 . . . . in³

The corresponding decimal values of x and volume are ...

... x ≈ 0.83 in

... v ≈ 1.16 in³

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