To solve this problem, let’s go through each step carefully.

1. Define the Variables:

   • The rectangular sheet is initially 3 inches by 8 inches.
   • We cut squares of side length $x$ from each corner. This makes the length of the box’s base $(8 - 2x)$ and the width of the base $(3 - 2x)$.
   • When the metal is folded up, the height of the box becomes $x$.

2. Set Up the Volume Formula:

   • The volume $V$ of an open-topped box is given by: $V = \text{length} \times \text{width} \times \text{height}$
   • Substituting for length, width, and height in terms of $x$: $V(x) = (8 - 2x)(3 - 2x)(x)$

3. Expand the Volume Expression:

   • Expanding $(8 - 2x)(3 - 2x)$ first: $(8 - 2x)(3 - 2x) = 24 - 16x - 6x + 4x^2 = 4x^2 - 22x + 24$
   • Then, multiply by $x$ for the volume: $V(x) = x(4x^2 - 22x + 24) = 4x^3 - 22x^2 + 24x$

4. Differentiate to Find Maximum Volume:

   • To find the maximum, take the derivative of $V(x)$ with respect to $x$ and set it to zero. $V'(x) = 12x^2 - 44x + 24$
   • Set $V'(x) = 0$ and solve for $x$: $12x^2 - 44x + 24 = 0$
   • Dividing by 4 to simplify: $3x^2 - 11x + 6 = 0$
   • Solving this quadratic equation using the quadratic formula: $x = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 3 \cdot 6}}{2 \cdot 3}$ $x = \frac{11 \pm \sqrt{121 - 72}}{6}$ $x = \frac{11 \pm \sqrt{49}}{6}$ $x = \frac{11 \pm 7}{6}$
     • Thus, $x = 3$ or $x = \frac{2}{3}$.

5. Determine Feasible Solution:

   • Since $x = 3$ would result in a width of zero (since $3 - 2(3) = 0$), it’s not feasible.
   • Therefore, $x = \frac{2}{3}$ is the only feasible solution.

6. Calculate the Maximum Volume:

   • Substitute $x = \frac{2}{3}$ into the volume formula: $V\left(\frac{2}{3}\right) = 4\left(\frac{2}{3}\right)^3 - 22\left(\frac{2}{3}\right)^2 + 24\left(\frac{2}{3}\right)$
   • Simplify each term and calculate for the exact volume.

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