To solve this optimization problem, let’s break it down step by step:

1. Define the Problem:

   • You have a rectangular piece of cardboard with dimensions $12 \times 20$ inches.
   • Squares of side length $x$ are cut out from each corner.
   • The resulting box has dimensions:
     • Length = $20 - 2x$,
     • Width = $12 - 2x$,
     • Height = $x$.

   The goal is to maximize the volume of the box.

2. Write the Volume Function: The volume $V$ of the box is given by: $V(x) = (20 - 2x)(12 - 2x)(x)$

   Simplify: $V(x) = x(240 - 40x - 24x + 4x^2)$ $V(x) = 4x^3 - 64x^2 + 240x$

3. Find the Critical Points: To maximize the volume, we find the derivative of $V(x)$ and set it to 0: $V'(x) = 12x^2 - 128x + 240$

   Solve $V'(x) = 0$ for $x$: $12x^2 - 128x + 240 = 0$

   Divide through by 4: $3x^2 - 32x + 60 = 0$

   Solve using the quadratic formula: $x = \frac{-(-32) \pm \sqrt{(-32)^2 - 4(3)(60)}}{2(3)}$ $x = \frac{32 \pm \sqrt{1024 - 720}}{6}$ $x = \frac{32 \pm \sqrt{304}}{6}$ Approximate $\sqrt{304}$: $x = \frac{32 \pm 17.44}{6}$

   Two solutions: $x = \frac{32 + 17.44}{6} \approx 8.24 \quad \text{(not valid, as it exceeds dimensions of the cardboard)},$ $x = \frac{32 - 17.44}{6} \approx 2.41$

4. Check the Validity and Dimensions: When $x \approx 2.41$:

   • Length = $20 - 2(2.41) \approx 15.18$,
   • Width = $12 - 2(2.41) \approx 7.18$,
   • Height = $x \approx 2.41$.

5. Confirm the Maximum: Check the volume at $x = 2.41$ and ensure it is the maximum using the second derivative or endpoints. You will find that this value indeed maximizes the volume.

Final Answer:

The dimensions of the box are approximately:

• Length: 15.18 inches,
• Width: 7.18 inches,
• Height: 2.41 inches.

Would you like detailed graphs, verification steps, or further clarifications?

5 Related Questions:

1. What happens to the volume if the cardboard dimensions are doubled?
2. Can the same formula be used for a square cardboard?
3. How do the results change if the cardboard is 12x24 inches instead of 12x20 inches?
4. How do you interpret the second derivative test in optimization problems like this?
5. What practical applications can be drawn from this type of problem?

Tip:

When solving optimization problems, always double-check boundary conditions to ensure your solution makes physical sense!