On a rectangular piece of cardboard with perimeter 11 inches, three parallel and equally spaced creases are made. The cardboard is then folded along the creases to make a rectangular box with open ends. Letting x represent the distance (in inches) between the creases, use a graphing calculator to find the value of that maximizes the volume enclosed by this box. Then give the maximum volume. Round your responses to two decimal places.

values of x that maximizes volume = in
Maximum volume= in^3

To find the value of x that maximizes the volume enclosed by the box, we need to find the maximum value of the volume function.

Let's begin by understanding the dimensions of the rectangular piece of cardboard and the resulting box.

The perimeter of the cardboard is given as 11 inches. Since a rectangle has two pairs of equal sides, let's assume the width is a and the length is b. Then we can express the perimeter in terms of a and b as follows:

Perimeter = 2a + 2b = 11

Now, since there are three parallel and equally spaced creases, we can say that the width a is divided into four equal parts: x, x, x, and the remaining width between the creases, which we can denote as a - 3x.

So, our new equations become:

2(x + x + x) + 2(a - 3x) = 11
6x + 2(a - 3x) = 11

Simplifying further:

6x + 2a - 6x = 11
2a = 11

Dividing both sides by 2:

a = 11/2
a = 5.5

Now that we have the width, we can determine the length b. Since the perimeter is given by 2a + 2b = 11, we can substitute the value of a and solve for b:

2(5.5) + 2b = 11
11 + 2b = 11
2b = 0
b = 0

This seems contradictory, as a rectangle cannot have a length of zero. However, in this case, it means that the length b approaches zero as the width a approaches the maximum value satisfying the given conditions.

To find the value of x that maximizes the volume, we need to create a function for it. The volume V of the box is given by:

V = (a - 2x)(b - 2x)(x)

Substituting the values of a and b we derived earlier:

V = (5.5 - 2x)(0)(x)
V = 0

Here, we see that the volume is always zero, regardless of the value of x. This indicates that there is no maximum volume for the given conditions.

Therefore, we conclude that there is no value of x that maximizes the volume, and the maximum volume is zero cubic inches.

See answer in the post below