Question

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The volume of an open top box is created by cutting a square from each corner of a 10 in. by 12 in. piece of paper. The situation is represented by the function V(x)=x(10−2x)(12−2x)
. Find the maximum volume of a box that can be created from the piece of paper. Round the answer to one decimal place.

(1 point)
cubic inches

Answer 1

To find the maximum volume of the box represented by the function \( V(x) = x(10 - 2x)(12 - 2x) \), we'll first need to simplify and analyze this function.

Let's start by expanding the function:

1. Start by expanding \( (10 - 2x)(12 - 2x) \):
\[
(10 - 2x)(12 - 2x) = 120 - 20x - 24x + 4x^2 = 120 - 44x + 4x^2
\]

2. Now substitute this back into the volume function:
\[
V(x) = x(120 - 44x + 4x^2) = 120x - 44x^2 + 4x^3
\]

The volume function is:
\[
V(x) = 4x^3 - 44x^2 + 120x
\]

Next, we'll find the critical points by taking the derivative of \( V(x) \) and setting it to zero:

3. Calculate the first derivative \( V'(x) \):
\[
V'(x) = 12x^2 - 88x + 120
\]

4. Set the derivative equal to zero:
\[
12x^2 - 88x + 120 = 0
\]

5. Simplify the quadratic equation by dividing by 4:
\[
3x^2 - 22x + 30 = 0
\]

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

Calculating the two possible critical points:

- \( x_1 = \frac{11 + \sqrt{31}}{3} \)
- \( x_2 = \frac{11 - \sqrt{31}}{3} \)

Now, we will find the approximate values:

Using \( \sqrt{31} \approx 5.57 \):
- \( x_1 \approx \frac{11 + 5.57}{3} \approx \frac{16.57}{3} \approx 5.19 \)
- \( x_2 \approx \frac{11 - 5.57}{3} \approx \frac{5.43}{3} \approx 1.81 \)

Since \( x \) must be between \( 0 \) and \( 5 \) because we cut squares from all corners:

Next, we need to evaluate \( V(x) \) at \( x \approx 1.81 \) since \( x_1 = 5.19 \) is outside the bounds (cut squares cannot exceed half the shorter side).

Now, calculate \( V(1.81) \):

\[
V(1.81) = 4(1.81)^3 - 44(1.81)^2 + 120(1.81)
\]
Calculating each term:
\[
(1.81)^3 \approx 5.92
\]
\[
(1.81)^2 \approx 3.28
\]
Hence,
\[
V(1.81) \approx 4(5.92) - 44(3.28) + 120(1.81)
\]
Computing the approximate values:
\[
\approx 23.68 - 144.32 + 217.2 \approx 96.56
\]

Thus, rounding to one decimal point, the maximum volume \( V\) of the open box is approximately \( \text{96.6 cubic inches} \).