Question
side of the square that is cut out has length x.
Complete parts (a) through (e).
38
38
38
38
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
enter your response here 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
enter your response here inches by
enter your response here 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
enter your response here.
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.)
Part 5
(e) For which of these values of x does a box exist if squares of length x are cut out and the tabs folded up?
A.
For neither of the values of x.
B.
For the smaller value of x.
C.
For all the values of x.
D.
For the larger value of x.
Complete parts (a) through (e).
38
38
38
38
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
enter your response here 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
enter your response here inches by
enter your response here 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
enter your response here.
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.)
Part 5
(e) For which of these values of x does a box exist if squares of length x are cut out and the tabs folded up?
A.
For neither of the values of x.
B.
For the smaller value of x.
C.
For all the values of x.
D.
For the larger value of x.
Answers
GPT-4o mini
To solve the problem related to constructing a box from a square piece of cardboard by cutting out squares of side length \( x \) from each corner, we will complete parts (a) through (e) step by step.
### **Part 1** (a)
**Write an expression for the height of the box that is constructed.**
The height of the box is equal to the side length of the square that is cut out:
\[
\text{Height of the box} = x \text{ inches}
\]
### **Part 2** (b)
**Write an expression for the dimensions of the base of the box that is constructed.**
If the original piece of cardboard is \( 38 \) inches on each side, cutting out squares of side \( x \) on each corner will reduce both the length and width of the base by \( 2x \) (since squares are cut from both sides):
\[
\text{Length of the base} = 38 - 2x \text{ inches}
\]
\[
\text{Width of the base} = 38 - 2x \text{ inches}
\]
So, the dimensions of the base of the constructed box are:
\[
(38 - 2x) \text{ inches by } (38 - 2x) \text{ inches}
\]
### **Part 3** (c)
**Use the formula \( V = lwh \) to find an equation that represents the volume of the box.**
Substituting the expressions for length, width, and height into the volume formula:
\[
V = (38 - 2x)(38 - 2x)(x)
\]
Simplifying:
\[
V = (38 - 2x)^2 \cdot x
\]
Thus, the volume of the box is:
\[
V = x(38 - 2x)^2
\]
### **Part 4** (d)
**Use the equation of the volume of the box to find the values of \( x \) that make \( V = 0 \).**
To solve \( V = 0 \):
\[
x(38 - 2x)^2 = 0
\]
This gives us two conditions:
1. \( x = 0 \)
2. \( (38 - 2x)^2 = 0 \) → \( 38 - 2x = 0 \) → \( 2x = 38 \) → \( x = 19 \)
So the values of \( x \) that make \( V = 0 \) are:
\[
x = 0, 19
\]
### **Part 5** (e)
**For which of these values of \( x \) does a box exist if squares of length \( x \) are cut out and the tabs folded up?**
- \( x = 0 \): Here the height is \( 0 \), thus a box does not exist.
- \( x = 19 \): Cutting out squares of this size means the base dimensions will be \( 38 - 2(19) = 0 \), which means no box can exist either.
Since for both values of \( x \), the volume ends up being zero or not creating a valid box, the correct answer is:
\[
\text{A. For neither of the values of } x.
\]
### **Final Summary**
- **(a)** Height: \( x \)
- **(b)** Dimensions of base: \( (38 - 2x) \text{ inches by } (38 - 2x) \text{ inches} \)
- **(c)** Volume: \( V = x(38 - 2x)^2 \)
- **(d)** Values of \( x \) for \( V = 0 \): \( 0, 19 \)
- **(e)** Box existence: \( A. \text{ For neither of the values of } x. \)
### **Part 1** (a)
**Write an expression for the height of the box that is constructed.**
The height of the box is equal to the side length of the square that is cut out:
\[
\text{Height of the box} = x \text{ inches}
\]
### **Part 2** (b)
**Write an expression for the dimensions of the base of the box that is constructed.**
If the original piece of cardboard is \( 38 \) inches on each side, cutting out squares of side \( x \) on each corner will reduce both the length and width of the base by \( 2x \) (since squares are cut from both sides):
\[
\text{Length of the base} = 38 - 2x \text{ inches}
\]
\[
\text{Width of the base} = 38 - 2x \text{ inches}
\]
So, the dimensions of the base of the constructed box are:
\[
(38 - 2x) \text{ inches by } (38 - 2x) \text{ inches}
\]
### **Part 3** (c)
**Use the formula \( V = lwh \) to find an equation that represents the volume of the box.**
Substituting the expressions for length, width, and height into the volume formula:
\[
V = (38 - 2x)(38 - 2x)(x)
\]
Simplifying:
\[
V = (38 - 2x)^2 \cdot x
\]
Thus, the volume of the box is:
\[
V = x(38 - 2x)^2
\]
### **Part 4** (d)
**Use the equation of the volume of the box to find the values of \( x \) that make \( V = 0 \).**
To solve \( V = 0 \):
\[
x(38 - 2x)^2 = 0
\]
This gives us two conditions:
1. \( x = 0 \)
2. \( (38 - 2x)^2 = 0 \) → \( 38 - 2x = 0 \) → \( 2x = 38 \) → \( x = 19 \)
So the values of \( x \) that make \( V = 0 \) are:
\[
x = 0, 19
\]
### **Part 5** (e)
**For which of these values of \( x \) does a box exist if squares of length \( x \) are cut out and the tabs folded up?**
- \( x = 0 \): Here the height is \( 0 \), thus a box does not exist.
- \( x = 19 \): Cutting out squares of this size means the base dimensions will be \( 38 - 2(19) = 0 \), which means no box can exist either.
Since for both values of \( x \), the volume ends up being zero or not creating a valid box, the correct answer is:
\[
\text{A. For neither of the values of } x.
\]
### **Final Summary**
- **(a)** Height: \( x \)
- **(b)** Dimensions of base: \( (38 - 2x) \text{ inches by } (38 - 2x) \text{ inches} \)
- **(c)** Volume: \( V = x(38 - 2x)^2 \)
- **(d)** Values of \( x \) for \( V = 0 \): \( 0, 19 \)
- **(e)** Box existence: \( A. \text{ For neither of the values of } x. \)