The critical points found are $x = \frac{4}{3}$ and $x = 4$. However, the maximum volume calculation results in zero when using $x = \frac{4}{3}$. This suggests that there may have been an issue with the substitution or calculation.

Let's manually re-evaluate the volume using the valid critical point $x = \frac{4}{3}$:

Volume Formula:

$V(x) = 4x^3 - 32x^2 + 64x$

Substituting $x = \frac{4}{3}$:

$V\left(\frac{4}{3}\right) = 4\left(\frac{4}{3}\right)^3 - 32\left(\frac{4}{3}\right)^2 + 64\left(\frac{4}{3}\right)$

I'll re-calculate this to ensure the correct maximum volume.The maximum volume of the box, when $x = \frac{4}{3}$ inches, is approximately 37.93 cubic inches.

This value corresponds to the dimensions:

• Length and Width of the base: $8 - 2\left(\frac{4}{3}\right) = \frac{20}{3}$ inches (approximately 6.67 inches)
• Height: $\frac{4}{3}$ inches

Would you like any further details or have any questions?

Related Questions:

1. How would the maximum volume change if the initial cardboard size was 10 inches by 10 inches?
2. What are the physical constraints that could limit the size of the squares cut from the corners?
3. Can you generalize the formula for a square cardboard of any size $s$?
4. What happens to the box's shape as $x$ approaches its minimum or maximum value?
5. How would the optimal $x$ change if we needed a closed box?

Tip:

Always double-check critical points in optimization problems to ensure they fall within the feasible domain of the problem.