Question
The volume of an open top box is created by cutting a square from each corner of a 10 inch by 12 inch 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 by identifying the relative maximum. Round the answer to one decimal place. (1 point) cubic inches
Answers
GPT-4o mini
To find the maximum volume of the box created by cutting squares of side length \( x \) from each corner of a 10 inch by 12 inch piece of paper, we can use the function given:
\[
V(x) = x(10 - 2x)(12 - 2x)
\]
First, expand this function.
1. **Expand the function**:
Begin by expanding \( (10 - 2x)(12 - 2x) \):
\[
(10 - 2x)(12 - 2x) = 120 - 20x - 24x + 4x^2 = 120 - 44x + 4x^2
\]
Now substitute back into \( V(x) \):
\[
V(x) = x(120 - 44x + 4x^2) = 120x - 44x^2 + 4x^3
\]
Thus, the volume function can be written as:
\[
V(x) = 4x^3 - 44x^2 + 120x
\]
2. **Find the derivative** \( V'(x) \) to locate the critical points:
\[
V'(x) = 12x^2 - 88x + 120
\]
3. **Set the derivative to zero to find critical points**:
\[
12x^2 - 88x + 120 = 0
\]
To simplify, divide the entire equation by 4:
\[
3x^2 - 22x + 30 = 0
\]
Now apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = 3, \quad b = -22, \quad c = 30
\]
\[
x = \frac{-(-22) \pm \sqrt{(-22)^2 - 4 \cdot 3 \cdot 30}}{2 \cdot 3}
\]
Calculate the discriminant:
\[
(-22)^2 - 4 \cdot 3 \cdot 30 = 484 - 360 = 124
\]
Now plug it into the quadratic formula:
\[
x = \frac{22 \pm \sqrt{124}}{6} = \frac{22 \pm 2\sqrt{31}}{6} = \frac{11 \pm \sqrt{31}}{3}
\]
Carrying out approximate calculations:
\[
\sqrt{31} \approx 5.57 \implies x \approx \frac{11 \pm 5.57}{3}
\]
This gives:
\[
x_1 \approx \frac{16.57}{3} \approx 5.19 \quad \text{and} \quad x_2 \approx \frac{5.43}{3} \approx 1.81
\]
Since \( x \) must be within the limits \( 0 < x < 5 \) (as cutting squares larger than that will not yield a box), only \( x \approx 1.81 \) is valid.
4. **To determine if this critical point is a maximum, evaluate the second derivative \( V''(x) \)**:
\[
V''(x) = 24x - 88
\]
Evaluate \( V''(1.81) \):
\[
V''(1.81) = 24(1.81) - 88 \approx 43.44 - 88 = -44.56
\]
Since \( V''(1.81) < 0 \), this confirms \( x = 1.81 \) is a relative maximum.
5. **Finally, calculate the maximum volume**:
\[
V(1.81) = 4(1.81)^3 - 44(1.81)^2 + 120(1.81)
\]
Perform each term:
\[
1.81^3 \approx 5.95 \quad \Rightarrow \quad 4 \cdot 5.95 \approx 23.8
\]
\[
1.81^2 \approx 3.28 \quad \Rightarrow \quad 44 \cdot 3.28 \approx 144.32
\]
\[
120 \cdot 1.81 \approx 217.2
\]
Therefore,
\[
V(1.81) \approx 23.8 - 144.32 + 217.2 \approx 96.68 \text{ cubic inches}
\]
After rounding to one decimal place, the maximum volume of the box that can be created from the piece of paper is approximately:
\[
\boxed{96.7} \text{ cubic inches}
\]
\[
V(x) = x(10 - 2x)(12 - 2x)
\]
First, expand this function.
1. **Expand the function**:
Begin by expanding \( (10 - 2x)(12 - 2x) \):
\[
(10 - 2x)(12 - 2x) = 120 - 20x - 24x + 4x^2 = 120 - 44x + 4x^2
\]
Now substitute back into \( V(x) \):
\[
V(x) = x(120 - 44x + 4x^2) = 120x - 44x^2 + 4x^3
\]
Thus, the volume function can be written as:
\[
V(x) = 4x^3 - 44x^2 + 120x
\]
2. **Find the derivative** \( V'(x) \) to locate the critical points:
\[
V'(x) = 12x^2 - 88x + 120
\]
3. **Set the derivative to zero to find critical points**:
\[
12x^2 - 88x + 120 = 0
\]
To simplify, divide the entire equation by 4:
\[
3x^2 - 22x + 30 = 0
\]
Now apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[
a = 3, \quad b = -22, \quad c = 30
\]
\[
x = \frac{-(-22) \pm \sqrt{(-22)^2 - 4 \cdot 3 \cdot 30}}{2 \cdot 3}
\]
Calculate the discriminant:
\[
(-22)^2 - 4 \cdot 3 \cdot 30 = 484 - 360 = 124
\]
Now plug it into the quadratic formula:
\[
x = \frac{22 \pm \sqrt{124}}{6} = \frac{22 \pm 2\sqrt{31}}{6} = \frac{11 \pm \sqrt{31}}{3}
\]
Carrying out approximate calculations:
\[
\sqrt{31} \approx 5.57 \implies x \approx \frac{11 \pm 5.57}{3}
\]
This gives:
\[
x_1 \approx \frac{16.57}{3} \approx 5.19 \quad \text{and} \quad x_2 \approx \frac{5.43}{3} \approx 1.81
\]
Since \( x \) must be within the limits \( 0 < x < 5 \) (as cutting squares larger than that will not yield a box), only \( x \approx 1.81 \) is valid.
4. **To determine if this critical point is a maximum, evaluate the second derivative \( V''(x) \)**:
\[
V''(x) = 24x - 88
\]
Evaluate \( V''(1.81) \):
\[
V''(1.81) = 24(1.81) - 88 \approx 43.44 - 88 = -44.56
\]
Since \( V''(1.81) < 0 \), this confirms \( x = 1.81 \) is a relative maximum.
5. **Finally, calculate the maximum volume**:
\[
V(1.81) = 4(1.81)^3 - 44(1.81)^2 + 120(1.81)
\]
Perform each term:
\[
1.81^3 \approx 5.95 \quad \Rightarrow \quad 4 \cdot 5.95 \approx 23.8
\]
\[
1.81^2 \approx 3.28 \quad \Rightarrow \quad 44 \cdot 3.28 \approx 144.32
\]
\[
120 \cdot 1.81 \approx 217.2
\]
Therefore,
\[
V(1.81) \approx 23.8 - 144.32 + 217.2 \approx 96.68 \text{ cubic inches}
\]
After rounding to one decimal place, the maximum volume of the box that can be created from the piece of paper is approximately:
\[
\boxed{96.7} \text{ cubic inches}
\]