Let's solve this step by step:

Part (a)

We have a rectangular piece of cardboard measuring $37 \, \text{ft} \times 20 \, \text{ft}$. Squares with sides of length $x$ are cut out from each corner, and the resulting piece of cardboard is folded into a box without a lid.

1. Volume of the box:

   • After cutting squares, the dimensions of the base of the box will be: $\text{Length} = 37 - 2x, \quad \text{Width} = 20 - 2x$
   • The height of the box will be $x$ (the height formed by folding the flaps).
   • Therefore, the volume $V$ of the box is: $V(x) = x \cdot (37 - 2x) \cdot (20 - 2x)$

2. Simplify the equation: Expand $V(x)$: $V(x) = x \cdot (740 - 74x - 40x + 4x^2)$ $V(x) = x \cdot (740 - 114x + 4x^2)$ $V(x) = 740x - 114x^2 + 4x^3$

3. Find the critical points: To maximize the volume, we take the derivative of $V(x)$ with respect to $x$ and set it to zero: $V'(x) = 740 - 228x + 12x^2$ Set $V'(x) = 0$: $12x^2 - 228x + 740 = 0$

4. Solve the quadratic equation: Divide through by 4 to simplify: $3x^2 - 57x + 185 = 0$ Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 3, b = -57, c = 185$ $x = \frac{-(-57) \pm \sqrt{(-57)^2 - 4(3)(185)}}{2(3)}$ $x = \frac{57 \pm \sqrt{3249 - 2220}}{6}$ $x = \frac{57 \pm \sqrt{1029}}{6}$ Approximate $\sqrt{1029} \approx 32.09$: $x = \frac{57 \pm 32.09}{6}$ $x = \frac{57 + 32.09}{6} \quad \text{or} \quad x = \frac{57 - 32.09}{6}$ $x \approx 14.51 \quad \text{or} \quad x \approx 4.14$

5. Choose the feasible solution: Since $x = 14.51$ would result in negative dimensions ($37 - 2x$ or $20 - 2x$), we choose $x = 4.14$.

6. Calculate the maximum volume: Substitute $x = 4.14$ into $V(x)$: $V(4.14) = 4.14 \cdot (37 - 2(4.14)) \cdot (20 - 2(4.14))$ $V(4.14) = 4.14 \cdot (37 - 8.28) \cdot (20 - 8.28)$ $V(4.14) = 4.14 \cdot 28.72 \cdot 11.72$ $V(4.14) \approx 1390.11 \, \text{ft}^3$

Part (b)

For a square cardboard with side $s$, the steps are similar:

1. Volume formula: $V(x) = x \cdot (s - 2x) \cdot (s - 2x)$ $V(x) = x(s - 2x)^2$ Expand and solve for $x$ to maximize $V(x)$.

Would you like detailed calculations for part (b) as well, or do you have questions about part (a)?

Expanding questions:

1. How does the choice of $x$ affect the volume of the box?
2. Why do we reject one of the solutions for $x$ from the quadratic equation?
3. Can you explain why the critical points are important for optimization problems?
4. How does the formula for part (b) differ from part (a)?
5. What other practical constraints might affect the choice of $x$?

Tip: Always check the feasibility of your solution by verifying the dimensions after solving for critical points.