To find the maximum volume of the box represented by the function \( V(x) = x(10 - 2x)(12 - 2x) \), we first need to expand this function and then find its critical points.
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Expand the volume function: \[ V(x) = x(10 - 2x)(12 - 2x) \] First, multiply \( (10 - 2x)(12 - 2x) \): \[ (10 - 2x)(12 - 2x) = 120 - 20x - 24x + 4x^2 = 120 - 44x + 4x^2 \] Now substitute this back into \( V(x) \): \[ V(x) = x(120 - 44x + 4x^2) = 120x - 44x^2 + 4x^3 \] Therefore, the volume function is: \[ V(x) = 4x^3 - 44x^2 + 120x \]
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Find the first derivative and critical points: We find the first derivative \( V'(x) \): \[ V'(x) = 12x^2 - 88x + 120 \] Set the derivative equal to zero to find critical points: \[ 12x^2 - 88x + 120 = 0 \] Dividing the entire equation by 4 simplifies this to: \[ 3x^2 - 22x + 30 = 0 \] We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{22 \pm \sqrt{(-22)^2 - 4 \cdot 3 \cdot 30}}{2 \cdot 3} = \frac{22 \pm \sqrt{484 - 360}}{6} = \frac{22 \pm \sqrt{124}}{6} \] Simplifying \( \sqrt{124} \): \[ \sqrt{124} = 2\sqrt{31} \] So: \[ x = \frac{22 \pm 2\sqrt{31}}{6} = \frac{11 \pm \sqrt{31}}{3} \]
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Determine valid critical points: Next, we must check the bounds for \( x \). The sides of the paper are 10 in. and 12 in., so: \[ 0 < x < 5 \quad \text{(due to the 10 in. side)} \] Evaluating \( \frac{11 - \sqrt{31}}{3} \) and \( \frac{11 + \sqrt{31}}{3} \): \[ \sqrt{31} \approx 5.57, \quad \text{thus}, \frac{11 - 5.57}{3} \approx 1.14 \quad \text{and} \quad \frac{11 + 5.57}{3} \approx 5.19 \] Only \( x = \frac{11 - \sqrt{31}}{3} \approx 1.14 \) is valid.
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Calculate maximum volume: Calculate \( V\left(\frac{11 - \sqrt{31}}{3}\right) \): Set \( x \approx 1.14 \): \[ V(1.14) = 4(1.14)^3 - 44(1.14)^2 + 120(1.14) \] Compute using a calculator: \[ V(1.14) \approx 4(1.481) - 44(1.2996) + 136.8 \approx 5.924 - 57.185 + 136.8 \approx 85.539 \text{ cubic inches} \] Rounding gives approximately: \[ \text{Maximum Volume} \approx 85.5 \text{ cubic inches} \]
So, the maximum volume of the box that can be created from the piece of paper is \( \boxed{85.5} \) cubic inches.
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