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

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.

GPT-4o mini · Answer

To find the maximum volume of the box formed by cutting squares of side length $x$ from each corner of a $10, in \times 12, in$ piece of paper, we use the given volume function:

$V (x) = x (10 - 2 x) (12 - 2 x)$

Step 1: Expand the volume function

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

$V (x) = x \cdot (10 - 2 x) \cdot (12 - 2 x)$

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

$(10 - 2 x) (12 - 2 x) = 120 - 20 x - 24 x + 4 x^{2} = 120 - 44 x + 4 x^{2}$

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

$V (x) = x (120 - 44 x + 4 x^{2}) = 120 x - 44 x^{2} + 4 x^{3}$

Thus, the volume function is:

$V (x) = 4 x^{3} - 44 x^{2} + 120 x$

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:

$V^{'} (x) = 12 x^{2} - 88 x + 120$

Set the derivative equal to zero:

$12 x^{2} - 88 x + 120 = 0$

Dividing the entire equation by 4 to simplify:

$3 x^{2} - 22 x + 30 = 0$

Step 3: Solve the quadratic equation

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

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

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:

$x_{1} = \frac{11 + \sqrt{31}}{3}$
$x_{2} = \frac{11 - \sqrt{31}}{3}$

Calculating numerical values (approximating):

$\sqrt{31} \approx 5.57 ⟹ x_{1} \approx \frac{11 + 5.57}{3} \approx \frac{16.57}{3} \approx 5.52 (out of bounds)$ $x_{2} \approx \frac{11 - 5.57}{3} \approx \frac{5.43}{3} \approx 1.81$

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:

$4 (1.81)^{3} \approx 4 \cdot 5.93 \approx 23.72$
$44 (1.81)^{2} \approx 44 \cdot 3.28 \approx 144.32$
$120 (1.81) \approx 120 \cdot 1.81 \approx 217.2$

Putting this all into the volume equation:

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

Step 6: Conclusion

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

$96.6, {in}^{3}$

Nanika Zoldyck · Answer

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.

GPT-4o mini · Answer

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.