An open box is to be formed from a square sheet of carboard (square is 10x10 cm) by cutting squares and then folding up the sides. (the squares cut off are just the corners as they are labeles as an x by x).

A) Find a function for the volume of the box. State the domain of your function.
B)Find the value of x that maximizes the volume.

v = x(10-2x)^2 = 4x(5-x)^2

You can see that v=0 when x=0 or 5, so somewhere in between, there is a maximum, since v>0 in that domain.

A bit of checking shows that x=5/3 makes v max.

To find a function for the volume of the box, you need to determine the dimensions of the box when the corners are cut off and the sides are folded up.

Let's assume that each side length of the square box formed after cutting the corners is y. When the corners are cut off, the dimensions of the resulting box will be (10-2x) by (10-2x) by y.

Thus, the function for the volume of the box can be expressed as V(x) = (10 - 2x) * (10 - 2x) * y.

Now, let's determine the domain of the function.

Since the size of the square cardboard is 10 cm x 10 cm, the maximum length that can be cut off from each side (x) is limited to half of the side length of the square cardboard, which is 10/2 = 5 cm. Therefore, the domain for x is from 0 to 5 cm.

Now, let's move on to determining the value of x that maximizes the volume.

To find the maximum volume, we need to find the critical points of the volume function, which occur when the derivative of the function is equal to zero or does not exist.

Taking the derivative of the volume function V(x) with respect to x:

V'(x) = [(10 - 2x)(10 - 2x) * y]'
= 2(10 - 2x) * (10 - 2x)' * y
= -8(10 - 2x) * y

Setting V'(x) = 0 and solving for x:

-8(10 - 2x) * y = 0
10 - 2x = 0
2x = 10
x = 5

Thus, x = 5 is the critical point that maximizes the volume of the box.

Please note that finding the y value that maximizes the volume requires more information, such as the specific dimensions or constraints of the problem.