To solve this problem, we need to find the value of $x$ that maximizes the volume of the open box created by cutting squares of side $x$ from each corner of an 18 cm by 18 cm square sheet of cardboard and folding up the sides.

Step 1: Define the Dimensions of the Box

1. After cutting out squares of side $x$ from each corner, the length and width of the base of the box will be $18 - 2x$.
2. The height of the box will be $x$, as that is the height formed by folding up the cut portions.

Step 2: Set Up the Volume Formula

The volume $V$ of a box is given by: $V = \text{length} \times \text{width} \times \text{height}$ For this box, we substitute: $V = (18 - 2x)(18 - 2x)(x)$ Expanding this, we get: $V = x(18 - 2x)^2$ Expanding further: $V = x(324 - 72x + 4x^2)$ Then distribute $x$ through the terms: $V = 4x^3 - 72x^2 + 324x$

Step 3: Find the Critical Points

To find the value of $x$ that maximizes the volume, we take the derivative of $V$ with respect to $x$ and set it equal to zero: $\frac{dV}{dx} = 12x^2 - 144x + 324$ Setting $\frac{dV}{dx} = 0$: $12x^2 - 144x + 324 = 0$ Divide by 12 to simplify: $x^2 - 12x + 27 = 0$ Now, solve this quadratic equation using the quadratic formula: $x = \frac{12 \pm \sqrt{(12)^2 - 4 \cdot 1 \cdot 27}}{2 \cdot 1}$ $x = \frac{12 \pm \sqrt{144 - 108}}{2}$ $x = \frac{12 \pm \sqrt{36}}{2}$ $x = \frac{12 \pm 6}{2}$ So, $x = 9$ or $x = 3$.

Step 4: Determine Which Value Maximizes the Volume

Since the cardboard is only 18 cm on a side, $x = 9$ would make the box nonexistent (since it would leave no base), so $x = 3$ is the feasible solution.

Step 5: Conclusion

The value of $x$ that maximizes the volume of the box is: $x = 3 \text{ cm}$

Would you like further details on the calculations, or have any questions?

Related Questions:

1. How would the volume formula change if the cardboard were rectangular instead of square?
2. What happens to the volume if we choose a different size of the square to cut out?
3. How can we verify that $x = 3$ maximizes the volume using the second derivative test?
4. What is the maximum volume that can be achieved with this value of $x$?
5. How would the solution change if the box had a lid?

Tip: When solving optimization problems, always check if critical points fall within the feasible range of the variable, as some values might not make sense physically.