Question

The volume of an open top box is created by cutting a square from each corner of a 10" x 12" piece of paper the situation is represented by the function VFX equals X 10 minus 2X 12 minus 2X find the maximum volume of a box that can be created from the piece of paper round the answer to one decimal place how many cubic units

Answer 1

Are you sure

Answer 2

What is the correct answer and why give me an explanation

Answer 3

To find the maximum volume of the open-top box created by cutting squares of side length \( x \) from each corner of a \( 10 \) inch by \( 12 \) inch piece of paper, we will start with the volume function you've provided, which is:

\[
V(x) = x(10 - 2x)(12 - 2x)
\]

The volume will be defined for \( x \) values where the dimensions remain positive, so \( x \) must satisfy the following conditions:

1. \( 10 - 2x > 0 \) which gives \( x < 5 \)
2. \( 12 - 2x > 0 \) which gives \( x < 6 \)

Thus, the feasible domain for \( x \) is \( 0 < x < 5 \).

Now, we can expand the volume function:

\[
V(x) = x(10 - 2x)(12 - 2x)
\]

First, let's simplify:

\[
V(x) = x \left( 120 - 20x - 24x + 4x^2 \right)
\]
\[
V(x) = x(120 - 44x + 4x^2)
\]
\[
V(x) = 120x - 44x^2 + 4x^3
\]

Next, we take the derivative and set it to zero to find the critical points:

\[
V'(x) = 120 - 88x + 12x^2
\]

Setting the derivative equal to zero:

\[
12x^2 - 88x + 120 = 0
\]

We can simplify the equation by dividing everything by 4:

\[
3x^2 - 22x + 30 = 0
\]

Now we can use the quadratic formula to solve for \( x \):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{22 \pm \sqrt{(-22)^2 - 4 \cdot 3 \cdot 30}}{2 \cdot 3}
\]
\[
= \frac{22 \pm \sqrt{484 - 360}}{6}
\]
\[
= \frac{22 \pm \sqrt{124}}{6}
\]
\[
= \frac{22 \pm 2\sqrt{31}}{6}
\]
\[
= \frac{11 \pm \sqrt{31}}{3}
\]

Calculating the approximate values of \( x \):

1. \( x_1 = \frac{11 + \sqrt{31}}{3} \approx \frac{11 + 5.57}{3} = \frac{16.57}{3} \approx 5.19 \) (not in domain)
2. \( x_2 = \frac{11 - \sqrt{31}}{3} \approx \frac{11 - 5.57}{3} = \frac{5.43}{3} \approx 1.81 \)

Next, we need to evaluate the volume at \( x \approx 1.81 \):

\[
V(1.81) = 1.81(10 - 2 \cdot 1.81)(12 - 2 \cdot 1.81)
\]
\[
= 1.81(10 - 3.62)(12 - 3.62)
\]
\[
= 1.81(6.38)(8.38)
\]
\[
\approx 1.81 \cdot 6.38 \cdot 8.38 \approx 1.81 \cdot 53.497 \approx 96.9 \text{ cubic inches}
\]

Thus, the maximum volume of the box, rounded to one decimal place, is:

\[
\boxed{96.9} \text{ cubic inches}
\]

Answer 4

Let's double-check the calculations step by step to ensure accuracy.

Given the dimensions of the paper, we want to find the volume of a box formed by cutting squares from each corner. The volume function based on cutting squares of side length \( x \) from a \( 10 \) inch by \( 12 \) inch sheet of paper is:

\[
V(x) = x(10 - 2x)(12 - 2x)
\]

Expanding this equation, we get:

1. Expand the inner terms:
\[
(10 - 2x)(12 - 2x) = 120 - 20x - 24x + 4x^2 = 120 - 44x + 4x^2
\]

2. Then, multiply by \( x \):
\[
V(x) = x(120 - 44x + 4x^2) = 120x - 44x^2 + 4x^3
\]

Next, we'll find the critical points by taking the derivative and setting it to zero:

\[
V'(x) = 120 - 88x + 12x^2
\]

Setting the derivative equal to zero:

\[
12x^2 - 88x + 120 = 0
\]

Dividing the whole equation by 4 simplifies it:

\[
3x^2 - 22x + 30 = 0
\]

Using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{22 \pm \sqrt{(-22)^2 - 4 \cdot 3 \cdot 30}}{2 \cdot 3}
\]
\[
= \frac{22 \pm \sqrt{484 - 360}}{6} = \frac{22 \pm \sqrt{124}}{6}
\]
\[
= \frac{22 \pm 2\sqrt{31}}{6} = \frac{11 \pm \sqrt{31}}{3}
\]

Now, calculating the approximate values:

1. \( \sqrt{31} \approx 5.57 \)
- \( x_1 = \frac{11 + 5.57}{3} \approx \frac{16.57}{3} \approx 5.19 \) (not valid, because \( x < 5 \))
- \( x_2 = \frac{11 - 5.57}{3} \approx \frac{5.43}{3} \approx 1.81 \) (valid)

Next, we'll calculate the volume at \( x \approx 1.81 \):

\[
V(1.81) = 1.81(10 - 2 \cdot 1.81)(12 - 2 \cdot 1.81)
\]
\[
= 1.81(10 - 3.62)(12 - 3.62)
\]
\[
= 1.81(6.38)(8.38)
\]

Now, let's calculate \( (6.38)(8.38) \):

\[
(6.38)(8.38) \approx 53.3394
\]

Now, calculate the volume:

\[
V(1.81) \approx 1.81 \times 53.3394 \approx 96.83
\]

Thus, rounding to one decimal place gives:

\[
\boxed{96.8} \text{ cubic inches}
\]

Thank you for your patience, and my earlier conclusion of \( 96.9 \) should indeed be corrected to \( 96.8 \). I appreciate your prompting for a re-evaluation!

Answer 5

Let's go through the problem step-by-step to ensure clarity and accuracy in finding the maximum volume of the box formed by cutting squares from each corner of a \( 10 \) inch by \( 12 \) inch piece of paper.

### Problem Summary
We are tasked with creating an open-top box by cutting out squares of side length \( x \) from each corner of a \( 10 \) inch by \( 12 \) inch piece of paper. The volume \( V \) of the resulting box can be expressed as:

\[
V(x) = x(10 - 2x)(12 - 2x)
\]

### Step 1: Volume Function

To derive the volume function:

1. **Start with:**
- The length will be reduced by \( 2x \) (from both ends), giving \( 12 - 2x \).
- The width will similarly be \( 10 - 2x \).

2. **Thus, the volume function is:**

\[
V(x) = x(10 - 2x)(12 - 2x)
\]

### Step 2: Expand the Volume Function

We can simplify this further:

1. **First, multiply out the inner terms:**

\[
(10 - 2x)(12 - 2x) = 120 - 20x - 24x + 4x^2 = 120 - 44x + 4x^2
\]

2. **Now multiply by \( x \):**

\[
V(x) = x(120 - 44x + 4x^2) = 120x - 44x^2 + 4x^3
\]

### Step 3: Find the Derivative

Next, we'll find the critical points by taking the derivative:

1. **Differentiate with respect to \( x \):**

\[
V'(x) = 120 - 88x + 12x^2
\]

2. **Set the derivative equal to zero:**

\[
12x^2 - 88x + 120 = 0
\]

### Step 4: Solve the Quadratic Equation

To solve for \( x \):

1. **Divide everything by 4 to simplify:**

\[
3x^2 - 22x + 30 = 0
\]

2. **Use the quadratic formula:**

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{22 \pm \sqrt{(-22)^2 - 4 \cdot 3 \cdot 30}}{2 \cdot 3}
\]

\[
= \frac{22 \pm \sqrt{484 - 360}}{6} = \frac{22 \pm \sqrt{124}}{6} = \frac{22 \pm 2\sqrt{31}}{6} = \frac{11 \pm \sqrt{31}}{3}
\]

3. **Approximate the roots:**

- \( \sqrt{31} \approx 5.57 \)
- \( x_1 = \frac{11 + 5.57}{3} \approx 5.19 \) (not valid since it exceeds \( 5 \))
- \( x_2 = \frac{11 - 5.57}{3} \approx 1.81 \) (valid since it is less than \( 5 \))

### Step 5: Calculate the Volume at Valid \( x \)

Now calculate the volume when \( x \approx 1.81 \):

1. **Substituting \( x = 1.81 \):**

\[
V(1.81) = 1.81(10 - 2(1.81))(12 - 2(1.81)) = 1.81(10 - 3.62)(12 - 3.62)
\]

2. **Calculate the values:**

\[
= 1.81(6.38)(8.38)
\]

- First, compute \( (6.38)(8.38) \approx 53.3394 \).

3. **Now find the volume:**

\[
V(1.81) \approx 1.81 \times 53.3394 \approx 96.81
\]

### Conclusion

Thus, the maximum volume, rounded to one decimal place, is:

\[
\boxed{96.8} \text{ cubic inches}
\]

### Explanation of Why This Works

- The method of using calculus (finding the derivative) allows us to determine where the volume function reaches its maximum.
- The physical constraints (which establish that \( x \) must be less than \( 5 \) inches) help us confirm that the value we ultimately calculate lies within valid bounds for the dimensions of the box.
- The method of evaluating the function at the valid critical point gives us the correct maximum volume for the constructed box.