Question
A box can be formed by cutting squares out of each corner of a piece of tin and folding the "tabs" up. Suppose the piece of tin is 8 inches by 8 inches and each side of the square that is cut out has length x.
Complete parts (a) through (e).
8
8
8
8
x
x
Question content area bottom
Part 1
(a) Write an expression for the height of the box that is constructed.
The height of the box is
x inches.
Part 2
(b) Write an expression for the dimensions of the base of the box that is constructed.
The dimensions of the base of the constructed box are
8 minus 2 x inches by
8 minus 2 x inches.
Part 3
(c) Use the formula Upper V equals lwh to find an equation that represents the volume of the box.
The volume of the box is
4 x cubed minus 32 x squared plus 64 x.
Part 4
(d) Use the equation of the volume of the box to find the values of x that make Upper V equals 0.
xequals
enter your response here (Use a comma to separate answers as needed.)
Complete parts (a) through (e).
8
8
8
8
x
x
Question content area bottom
Part 1
(a) Write an expression for the height of the box that is constructed.
The height of the box is
x inches.
Part 2
(b) Write an expression for the dimensions of the base of the box that is constructed.
The dimensions of the base of the constructed box are
8 minus 2 x inches by
8 minus 2 x inches.
Part 3
(c) Use the formula Upper V equals lwh to find an equation that represents the volume of the box.
The volume of the box is
4 x cubed minus 32 x squared plus 64 x.
Part 4
(d) Use the equation of the volume of the box to find the values of x that make Upper V equals 0.
xequals
enter your response here (Use a comma to separate answers as needed.)
Answers
GPT-4o mini
To find the volume and the values of \( x \) that make \( V = 0 \), let’s start by following the steps outlined above.
### Part 1
**(a)** The height of the box after cutting out squares of side \( x \) inches from each corner and folding up the tabs will be:
- **Height**: \( x \) inches.
### Part 2
**(b)** The dimensions of the base of the box will change as the squares are cut out. Each side of the square cut at the corner reduces the corresponding dimension by 2 times \( x \) (one from each side).
- **Base Dimensions**: \( (8 - 2x) \) inches by \( (8 - 2x) \) inches.
### Part 3
**(c)** The volume \( V \) of the box can be expressed as a function of its dimensions:
\[
V = \text{Length} \times \text{Width} \times \text{Height} = (8 - 2x)(8 - 2x)(x)
\]
This simplifies to:
\[
V = (8 - 2x)^2 \cdot x
\]
Now, we can expand:
\[
(8 - 2x)(8 - 2x) = 64 - 32x + 4x^2
\]
So then, multiplying by \( x \):
\[
V = x(64 - 32x + 4x^2) = 4x^3 - 32x^2 + 64x
\]
### Part 4
**(d)** To find the values of \( x \) that make \( V = 0 \), we can set the volume equation equal to zero:
\[
4x^3 - 32x^2 + 64x = 0
\]
We can factor this out:
\[
4x(x^2 - 8x + 16) = 0
\]
This gives us:
\[
4x = 0 \quad \text{or} \quad x^2 - 8x + 16 = 0
\]
From \( 4x = 0 \), we have \( x = 0 \).
For the quadratic part:
\[
x^2 - 8x + 16 = 0 \implies (x - 4)^2 = 0 \implies x = 4.
\]
Thus, the values of \( x \) that make \( V = 0 \) are:
\[
x = 0, 4.
\]
### Answer
Therefore, the values of \( x \) are \( \boxed{0, 4} \).
### Part 1
**(a)** The height of the box after cutting out squares of side \( x \) inches from each corner and folding up the tabs will be:
- **Height**: \( x \) inches.
### Part 2
**(b)** The dimensions of the base of the box will change as the squares are cut out. Each side of the square cut at the corner reduces the corresponding dimension by 2 times \( x \) (one from each side).
- **Base Dimensions**: \( (8 - 2x) \) inches by \( (8 - 2x) \) inches.
### Part 3
**(c)** The volume \( V \) of the box can be expressed as a function of its dimensions:
\[
V = \text{Length} \times \text{Width} \times \text{Height} = (8 - 2x)(8 - 2x)(x)
\]
This simplifies to:
\[
V = (8 - 2x)^2 \cdot x
\]
Now, we can expand:
\[
(8 - 2x)(8 - 2x) = 64 - 32x + 4x^2
\]
So then, multiplying by \( x \):
\[
V = x(64 - 32x + 4x^2) = 4x^3 - 32x^2 + 64x
\]
### Part 4
**(d)** To find the values of \( x \) that make \( V = 0 \), we can set the volume equation equal to zero:
\[
4x^3 - 32x^2 + 64x = 0
\]
We can factor this out:
\[
4x(x^2 - 8x + 16) = 0
\]
This gives us:
\[
4x = 0 \quad \text{or} \quad x^2 - 8x + 16 = 0
\]
From \( 4x = 0 \), we have \( x = 0 \).
For the quadratic part:
\[
x^2 - 8x + 16 = 0 \implies (x - 4)^2 = 0 \implies x = 4.
\]
Thus, the values of \( x \) that make \( V = 0 \) are:
\[
x = 0, 4.
\]
### Answer
Therefore, the values of \( x \) are \( \boxed{0, 4} \).