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)

Question

. 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 box formed by cutting squares of side length \( x \) from each corner of a \( 10 , \text{in} \times 12 , \text{in} \) piece of paper, we use the given volume function:

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

Step 1: Expand the volume function

First, we need to expand the function \( V(x) \):

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

Calculating \( (10 - 2x)(12 - 2x) \):

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

Now, substituting this back into \( V(x) \):

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

Thus, the volume function is:

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

Step 2: Find the critical points

To maximize the volume, we need to find the critical points by taking the derivative and setting it to zero:

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

Set the derivative equal to zero:

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

Dividing the entire equation by 4 to simplify:

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

Step 3: Solve the quadratic equation

To solve for \( x \), we can use the quadratic formula where \( a = 3, b = -22, c = 30 \):

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

Calculating the discriminant:

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

Now substituting back into the formula:

\[ x = \frac{22 \pm \sqrt{124}}{6} \]

Calculating \( \sqrt{124} = 2\sqrt{31} \):

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

Step 4: Evaluate the function at the critical points

We only consider the solutions that fall within the constraints \( 0 < x < 5 \) (the maximum cut is half the shorter side, which is 5 inches).

Calculating the two critical points:

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

Calculating numerical values (approximating):

\[ \sqrt{31} \approx 5.57 \implies x_1 \approx \frac{11 + 5.57}{3} \approx \frac{16.57}{3} \approx 5.52 \text{ (out of bounds)} \] \[ x_2 \approx \frac{11 - 5.57}{3} \approx \frac{5.43}{3} \approx 1.81 \]

Step 5: Evaluate the volume function at \( x = x_2 \)

Substituting \( x = 1.81 \) into the volume function:

\[ V(1.81) = 4(1.81)^3 - 44(1.81)^2 + 120(1.81) \]

Calculating each term:

1. \( 4(1.81)^3 \approx 4 \cdot 5.93 \approx 23.72 \)
2. \( 44(1.81)^2 \approx 44 \cdot 3.28 \approx 144.32 \)
3. \( 120(1.81) \approx 120 \cdot 1.81 \approx 217.2 \)

Putting this all into the volume equation:

\[ V(1.81) \approx 23.72 - 144.32 + 217.2 \approx 96.6 , \text{in}^3 \]

Step 6: Conclusion

Thus, the maximum volume of the box that can be created from the piece of paper is:

\[ \boxed{96.6} , \text{in}^3 \]

Answer 2

Francisco’s teacher asked him to describe the number of relative extrema a polynomial of degree five could have. Francisco answered, “The polynomial will have exactly five relative extrema.” Write 1–2 sentences discussing why Francisco’s answer is incorrect.

Answer 3

Francisco's answer is incorrect because a polynomial of degree five can have at most four relative extrema, not five. The number of relative extrema is determined by the degree of the polynomial minus one, meaning a degree \( n \) polynomial can have at most \( n - 1 \) relative extrema. In this case, with a degree of five, the maximum number of relative extrema is four.