A piece of cardboard is 2.4 times as long as it is wide. It is to be made into a box with an open top by cutting 3-inch squares from each comer and folding up the sides. Let x represent the width (in inches) of the original piece of cardboard Answer the following questions 3
a) Represent the length of the original piece of cardboard in terms of x. Length 2.4x in. (Use integers or decimals for any numbers in the expression.)
b) Give the restrictions on x. What will be the dimensions of the bottom rectangular base of the box? The restriction on x will be x>6. (Type an inequality.) The length will be (2.4x-6)in. and the width will be (x-6)in (Type expressions using x as the variable. Use integers or decimals for any numbers in the expressions.)
c) Determine a function V that represents the volume of the box in terms of x. V 7.2x61.2x108in (Simplify your answer. Use integers or decimals for any numbers in the expression.)
d) For what dimensions of the bottom of the box will the volume be 520 in.3? The length will be 25.03] in and the width will be [693] in. Round to the nearest tenth as needed.)
e) Find the values of x if such a box is to have a volume between 600 and 800 in.3 Between which two values must x be in order to produce this range of volumes? (Use a comma to separate answers as needed. Round to the nearest tenth as needed.)

The restriction on x is x > 6
The expression for volume is [tex]\mathbf{V = 7.2x^2 -61.2x +108}[/tex].
The dimension of the box for a volume of 520 is: 25.03 by 6.93 by 3
If the box is to have a volume between 600 and 800, the value of x would be between 13.54 and 14.93

(a) The dimension of the box

Let:

The width of the cardboard be x.

So, the length of the cardboard is: 2.4x

When 3 inches is removed, the dimension of the box is:

[tex]\mathbf{L = 2.4x- 6}[/tex] --- length

[tex]\mathbf{W = x - 6}[/tex] ---- width

[tex]\mathbf{H = 3}[/tex] --- height

(b) The restriction on x

When 3 inches is removed, it means that a total of 6 inches will be removed from either sides.

Hence, the value of x must be greater than 6.

So, the restriction on x is:

[tex]\mathbf{x > 6}[/tex]

(c) Function that represents volume.

The volume (V) of a box is:

[tex]\mathbf{V = L \times W \times H}[/tex]

So, we have:

[tex]\mathbf{V = (2.4x - 6) \times (x - 6) \times 3}[/tex]

Simplify

[tex]\mathbf{V = (7.2x - 18) \times (x - 6)}[/tex]

Expand

[tex]\mathbf{V = 7.2x^2 - 43.2x - 18x +108}[/tex]

[tex]\mathbf{V = 7.2x^2 -61.2x +108}[/tex]

(d) The dimension, when the volume is 520

This means that, V = 520

So, we have:

[tex]\mathbf{ 7.2x^2 -61.2x +108 = 520}[/tex]

Collect like terms

[tex]\mathbf{ 7.2x^2 -61.2x +108 - 520 = 0}[/tex]

[tex]\mathbf{ 7.2x^2 -61.2x -412 = 0}[/tex]

Using a calculator, we have:

[tex]\mathbf{x = 12.93\ or\ x = -4.43}[/tex]

Recall that: [tex]\mathbf{x > 6}[/tex]

So, the value of x is:

[tex]\mathbf{x = 12.93}[/tex]

Substitute 12.93 for x in [tex]\mathbf{L = 2.4x- 6}[/tex] and [tex]\mathbf{W = x - 6}[/tex]

So, we have:

[tex]\mathbf{L = 2.4 \times 12.93 - 6 = 25.03}[/tex]

[tex]\mathbf{W = 12.93 - 6 = 6.93}[/tex]

So, the dimension of the box is: 25.03 by 6.93 by 3

(e) The value of x, when the volume is between 600 and 800

This means that, V = 600 and V = 800

When V = 600, we have:

[tex]\mathbf{ 7.2x^2 -61.2x +108 = 600}[/tex]

Collect like terms

[tex]\mathbf{ 7.2x^2 -61.2x +108 - 600 = 0}[/tex]

[tex]\mathbf{ 7.2x^2 -61.2x -492 = 0}[/tex]

Using a calculator, we have:

[tex]\mathbf{x = 13.54\ or\ x = -5.04}[/tex]

Recall that: [tex]\mathbf{x > 6}[/tex]

So, the value of x is:

[tex]\mathbf{x = 13.54}[/tex]

When V = 800, we have:

[tex]\mathbf{ 7.2x^2 -61.2x +108 = 800}[/tex]

Collect like terms

[tex]\mathbf{ 7.2x^2 -61.2x +108 - 800 = 0}[/tex]

[tex]\mathbf{ 7.2x^2 -61.2x -692 = 0}[/tex]

Using a calculator, we have:

[tex]\mathbf{x = 14.93\ or\ x = -6.45}[/tex]

Recall that: [tex]\mathbf{x > 6}[/tex]

So, the value of x is:

[tex]\mathbf{x = 14.93}[/tex]

If such a box is to have a volume between 600 and 800, the value of x would be between 13.54 and 14.93