To find the largest volume of the open box that can be constructed, we use optimization. Here's the step-by-step process:

---

1. Define the problem mathematically:

Let $x$ be the length of the side of the square cut out from each corner.

• The dimensions of the resulting box:

  • Base: $(3 - 2x) \times (3 - 2x)$, since two squares of size $x$ are cut out from each side of the 3 ft square cardboard.
  • Height: $x$, since the sides are bent up to form the height.

• Volume (V) of the box: $V = \text{Base Area} \times \text{Height} = (3 - 2x)^2 \cdot x$

---

2. Express the volume equation:

Expanding $(3 - 2x)^2$: $(3 - 2x)^2 = 9 - 12x + 4x^2$

Thus, the volume equation becomes: $V(x) = x(9 - 12x + 4x^2) = 9x - 12x^2 + 4x^3$

---

3. Find the critical points (maximize $V(x)$):

Differentiate $V(x)$ with respect to $x$: $V'(x) = 9 - 24x + 12x^2$

Set $V'(x) = 0$: $12x^2 - 24x + 9 = 0$

Simplify: $4x^2 - 8x + 3 = 0$

Solve using the quadratic formula: $x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(4)(3)}}{2(4)}$ $x = \frac{8 \pm \sqrt{64 - 48}}{8}$ $x = \frac{8 \pm \sqrt{16}}{8}$ $x = \frac{8 \pm 4}{8}$ $x = \frac{12}{8} = 1.5 \quad \text{or} \quad x = \frac{4}{8} = 0.5$

---

4. Analyze critical points:

The feasible domain for $x$ is $0 < x < 1.5$, since $x = 1.5$ would make the base dimensions zero.

Calculate $V(x)$ at $x = 0.5$ and $x = 1.5$, and also analyze the endpoints.

1. At $x = 0.5$: $V(0.5) = 9(0.5) - 12(0.5)^2 + 4(0.5)^3 = 4.5 - 3 + 0.5 = 2 \, \text{cubic feet}$

2. At $x = 1.5$: Not feasible because it leads to zero dimensions for the base.

Check the endpoint $x \to 0$ (volume approaches zero as $x$ diminishes).

---

5. Conclusion:

The largest volume occurs at $x = 0.5$ and the volume is: $\boxed{2 \, \text{cubic feet}}$

---

Do you want further details or clarifications? Here are related questions to expand understanding:

1. How is the domain of $x$ determined in optimization problems like this?
2. What if the dimensions of the cardboard were different?
3. How would the problem change if the box had a closed top?
4. Can you graph $V(x)$ to visualize the maximum?
5. How does calculus help solve real-world optimization problems?

Tip: Always verify your critical points by checking the feasible domain and confirming maxima/minima with second derivative tests or endpoint analysis.