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 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

MrRoyal MrRoyal · Answer 2 · 2021-10-12T09:24:29+02:00

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

The value of x that maximizes the volume is: 0.83
The maximum volume is 1.16

The given parameter is:

[tex]\mathbf{Perimeter = 10}[/tex]

Let the height of the cardboard be h.

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

[tex]\mathbf{Perimeter = x + x + x + x+x + x + x + x + h + h}[/tex]

[tex]\mathbf{Perimeter = 8x + 2h}[/tex]

Substitute [tex]\mathbf{Perimeter = 10}[/tex]

[tex]\mathbf{8x + 2h = 10}[/tex]

Divide through by 2

[tex]\mathbf{4x + h = 5}[/tex]

Make h the subject

[tex]\mathbf{h = 5 -4x}[/tex]

The volume of the cardboard is then calculated as:

[tex]\mathbf{V = x^2h}[/tex]

Substitute [tex]\mathbf{h = 5 -4x}[/tex]

[tex]\mathbf{V = x^2(5 - 4x)}[/tex]

Expand

[tex]\mathbf{V = 5x^2- 4x^3}[/tex]

Differentiate

[tex]\mathbf{V' = 10x- 12x^2}[/tex]

Set to 0

[tex]\mathbf{10x- 12x^2 = 0}[/tex]

Rewrite as:

[tex]\mathbf{- 12x^2 = -10x}[/tex]

Divide through by -x

[tex]\mathbf{12x = 10}[/tex]

Divide through by 12

[tex]\mathbf{x = \frac{10}{12}}[/tex]

[tex]\mathbf{x = \frac{5}{6}}[/tex]

[tex]\mathbf{x = 0.83}[/tex]

Recall that:

[tex]\mathbf{V = x^2(5 - 4x)}[/tex]

So, we have:

[tex]\mathbf{V = (\frac 56)^2 \times (5 - 4 \times \frac 56)}[/tex]

[tex]\mathbf{V = \frac{25}{36} \times (5 - \frac{10}{3})}[/tex]

Take LCM

[tex]\mathbf{V = \frac{25}{36} \times \frac{15 - 10}{3}}[/tex]

[tex]\mathbf{V = \frac{25}{36} \times \frac{5}{3}}[/tex]

[tex]\mathbf{V = \frac{125}{108}}[/tex]

[tex]\mathbf{V = 1.16}[/tex]

Hence,

The value of x that maximizes the volume is: 0.83
The maximum volume is 1.16