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:
- Start by multiplying \( (10 - 2x)(12 - 2x) \):
\[ (10 - 2x)(12 - 2x) = 120 - 20x - 24x + 4x^2 = 120 - 44x + 4x^2 \]
- 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 \(x = 5\):
- At \(x = 0\):
\[ V(0) = 0(10 - 0)(12 - 0) = 0 \]
- At \(x = 5\):
\[ V(5) = 5(10 - 2 \cdot 5)(12 - 2 \cdot 5) = 5(0)(2) = 0 \]
- 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:
\[ \boxed{96.8} \text{ cubic inches.} \]
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