Question

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.

Answer 1

To find the maximum volume of the open-top box created by cutting squares of size \(x\) from each corner of a 10 inch by 12 inch piece of paper, we need to analyze the volume function given by:

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

First, we will expand the function:

1. Start by multiplying \( (10 - 2x)(12 - 2x) \):

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

2. Now substitute this result back into the volume function:

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

Next, we differentiate \(V(x)\) with respect to \(x\) to find the critical points:

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

Now, set the derivative equal to zero:

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

To simplify further, we can divide the entire equation by 4:

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

Next, we apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -22\), and \(c = 30\):

\[
b^2 - 4ac = (-22)^2 - 4 \cdot 3 \cdot 30 = 484 - 360 = 124
\]

Now, apply the quadratic formula:

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

Calculating the approximate values:

- \(\sqrt{31} \approx 5.57\)

Thus,

\[
x \approx \frac{11 + 5.57}{3} \approx \frac{16.57}{3} \approx 5.19 \quad (This is not valid since it exceeds \(\frac{10}{2} = 5\))
\]

\[
x \approx \frac{11 - 5.57}{3} \approx \frac{5.43}{3} \approx 1.81
\]

Now, we have a valid critical point \(x \approx 1.81\).

Next, we will check the volume at this critical point and also at the endpoints \(x = 0\) and \(x = 5\):

1. **At \(x = 0\)**:

\[
V(0) = 0(10 - 0)(12 - 0) = 0
\]

2. **At \(x = 5\)**:

\[
V(5) = 5(10 - 2 \cdot 5)(12 - 2 \cdot 5) = 5(0)(2) = 0
\]

3. **At \(x \approx 1.81\)**:

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

Calculating \(6.38 \cdot 8.38\):

\[
6.38 \cdot 8.38 \approx 53.39
\]

Then:

\[
V(1.81) \approx 1.81 \cdot 53.39 \approx 96.79
\]

Finally, rounding to one decimal place, the maximum volume of the box is:

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