To find the maximum volume of an open-top box made by cutting out congruent squares from each corner of a rectangular piece of tin, we’ll go through the following steps:

1. Define Variables and Set Up the Volume Equation:

   Let:

   • $x$ = side length of each square cut from the corners.
   • When squares are cut out, the box will have new dimensions of $(60 - 2x)$ inches in length and $(32 - 2x)$ inches in width, and the height of the box will be $x$ inches (from folding up the edges).

2. Volume of the Box:

   The volume $V$ of the box, given by $V = \text{length} \times \text{width} \times \text{height}$, can be written as: $V(x) = (60 - 2x)(32 - 2x)(x)$ Expanding this equation will help simplify our calculations.

3. Expanding the Volume Expression:

   Expanding $V(x) = (60 - 2x)(32 - 2x)x$: $V(x) = x (1920 - 120x - 64x + 4x^2)$ Simplifying the terms inside the parentheses: $V(x) = x (1920 - 184x + 4x^2)$ Expanding further: $V(x) = 1920x - 184x^2 + 4x^3$

   Thus, our volume function is: $V(x) = 4x^3 - 184x^2 + 1920x$

4. Find the Maximum Volume:

   To maximize the volume, we need to find the derivative of $V(x)$ with respect to $x$ and set it to zero: $V'(x) = 12x^2 - 368x + 1920$ Setting $V'(x) = 0$: $12x^2 - 368x + 1920 = 0$ Dividing through by 4: $3x^2 - 92x + 480 = 0$ Solving this quadratic equation using the quadratic formula: $x = \frac{92 \pm \sqrt{(-92)^2 - 4 \cdot 3 \cdot 480}}{2 \cdot 3}$ Simplifying: $x = \frac{92 \pm \sqrt{8464 - 5760}}{6}$ $x = \frac{92 \pm \sqrt{2704}}{6}$ $x = \frac{92 \pm 52}{6}$ The two solutions are: $x = \frac{144}{6} = 24 \quad \text{and} \quad x = \frac{40}{6} \approx 6.67$

   Since $x = 24$ would make the length and width negative, it is not feasible. Therefore, $x \approx 6.67$ inches is the only valid solution.

5. Calculate the Maximum Volume:

   Substitute $x = 6.67$ back into the volume formula: $V(6.67) = 4(6.67)^3 - 184(6.67)^2 + 1920(6.67)$ Calculating this: $V(6.67) \approx 4181.4 \text{ cubic inches}$

Diagram:

To illustrate, draw a rectangle labeled 60 in by 32 in, and mark congruent squares of side length $x$ being cut out from each corner. After folding up, label the dimensions as $(60 - 2x)$ by $(32 - 2x)$ with height $x$.

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Would you like more details or have any questions?

1. How is the volume affected if the dimensions of the tin change?
2. Why do we set the derivative of the volume to zero for maximization?
3. What does the constraint on $x$ tell us about the physical construction?
4. How would this change if the box had a lid?
5. What other optimization problems are similar in approach?

Tip: Solving for maximum or minimum values often involves taking derivatives and setting them to zero to find critical points.