help please !!!!!

Allie has a piece of construction paper that she wants to use to make an open rectangular prism. she will cut a square with side length x from each corner of the paper, so the length and width is decreased by 2x as shown in the diagram.

a.) write a function that models the volume of the rectangular prism
b.) graph the function and identify a reasonable domain
c.) what do the X intercepts of the graph mean in this context
d.) if Allie wants to maximize the volume of the Box what is the side length of the squares that should be cut from each corner of the piece of construction paper? explain

The function that models the volume is [tex]V(x) = 2x\cdot (11 - 2x)\cdot (8 - 2x)[/tex].
The reasonable domain is: (0, 4) u (4, 5.5)
The x-intercepts are the points where the paper is still flat
The lengths of the square that maximizes the volume of the box are 1.5 and 4.8

When the length and width of the paper is decreased, the dimension of the prism becomes

[tex]Length = 11-2x[/tex]

[tex]Width = 8 - 2x[/tex]

[tex]Height = 2x[/tex]

(a) The function

The function that represents the volume is:

[tex]V(x) = Length \times Width \times Height[/tex]

So, we have:

[tex]V(x) = (11 - 2x) \times (8 - 2x) \times 2x[/tex]

Rewrite as:

[tex]V(x) = 2x\cdot (11 - 2x)\cdot (8 - 2x)[/tex]

(b) The graph and the domain

See attachment for the graph of [tex]V(x) = 2x\cdot (11 - 2x)\cdot (8 - 2x)[/tex]

From the graph, we have the x-intercepts to be 0, 4 and 5.5

At these points, the volume of the prism is 0.

So, the reasonable domain is: (0, 4) u (4, 5.5)

(c) The interpretation of the x-intercepts

The x-intercepts are the points where the paper is still flat (i.e. the prism has no height or no width or no length)

(d) Dimension that maximizes the volume

We have:

[tex]V(x) = 2x\cdot (11 - 2x)\cdot (8 - 2x)[/tex]

Expand

[tex]V(x) = 2x \cdot (88 - 22x - 16x + 4x^2)[/tex]

[tex]V(x) = 2x \cdot (88 -38x + 4x^2)[/tex]

Open bracket

[tex]V(x) = 176x -76x^2 + 8x^3[/tex]

Differentiate

[tex]V'(x) = 176 -152x + 24x^2[/tex]

Set to 0

[tex]176 -152x + 24x^2=0[/tex]

Rewrite as:

[tex]24x^2 -152x + 176=0[/tex]

Divide through by 8

[tex]3x^2 -19x +22 =0[/tex]

Using a graphing calculator, the value of x is approximately:

[tex]x = 1.5[/tex] and [tex]x = 4.8[/tex]

Hence, the lengths of the square that maximizes the volume of the box are 1.5 and 4.8