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.
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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:
\[ V(x) = 4x^3 - 44x^2 + 120x \]
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Find the derivative \( V'(x) \) to locate the critical points:
\[ V'(x) = 12x^2 - 88x + 120 \]
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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.
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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.
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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:
\[ \boxed{96.7} \text{ cubic inches} \]
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