To find the maximum volume of the box created by cutting squares from each corner of a \(10\) inch by \(12\) inch piece of paper, given the function for volume \(V(X) = X(10 - 2X)(12 - 2X)\), we will first simplify this expression.
The volume function is represented as:
\[ V(X) = X(10 - 2X)(12 - 2X) \]
Let's simplify it step-by-step.
- Expand \( (10 - 2X)(12 - 2X) \):
\[ (10 - 2X)(12 - 2X) = 120 - 20X - 24X + 4X^2 = 120 - 44X + 4X^2 \]
- Now multiply by \(X\):
\[ V(X) = X(120 - 44X + 4X^2) = 120X - 44X^2 + 4X^3 \]
So, we have:
\[ V(X) = 4X^3 - 44X^2 + 120X \]
- To find the maximum volume, we need to take the derivative \(V'(X)\) and set it to zero:
\[ V'(X) = 12X^2 - 88X + 120 \]
Set the derivative to zero:
\[ 12X^2 - 88X + 120 = 0 \]
- Now, use the quadratic formula to solve for \(X\):
The quadratic formula is:
\[ X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 12\), \(b = -88\), and \(c = 120\):
\[ b^2 - 4ac = (-88)^2 - 4 \cdot 12 \cdot 120 = 7744 - 5760 = 1984 \]
Calculating \(X\):
\[ X = \frac{88 \pm \sqrt{1984}}{24} \]
Calculating \(\sqrt{1984} \approx 44.53\):
\[ X = \frac{88 \pm 44.53}{24} \]
This gives us two potential values for \(X\):
- \(X = \frac{132.53}{24} \approx 5.52\)
- \(X = \frac{43.47}{24} \approx 1.81\)
Since \(X\) must be less than half of the shorter side of the paper (6 inches), we disregard \(5.52\), leaving \(X \approx 1.81\).
- Now, we calculate the volume using \(X \approx 1.81\):
\[ V(1.81) = 4(1.81)^3 - 44(1.81)^2 + 120(1.81) \]
Calculating each part:
\[ (1.81)^3 \approx 5.935, (1.81)^2 \approx 3.276, \]
So,
\[ V(1.81) = 4(5.935) - 44(3.276) + 120(1.81) \] \[ = 23.74 - 143.44 + 217.2 \] \[ = 97.5 \]
Hence, rounding to one decimal place, the maximum volume of the box that can be created is:
\[ \boxed{97.5} \text{ cubic inches.} \]
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