To solve this problem, we need to find the dimensions of a box with an open top that minimizes the surface area, given a volume constraint. Let's break it down step-by-step.

Step 1: Set Up the Problem

Let:

• $x$ = the length of one side of the square base of the box
• $h$ = the height of the box

The volume $V$ of the box is given by: $V = x^2 \cdot h$ We know that $V = 97556 \, \text{cm}^3$. Therefore: $x^2 \cdot h = 97556$

From this, we can solve for $h$ in terms of $x$: $h = \frac{97556}{x^2}$

Step 2: Find the Surface Area in Terms of $x$

Since the box has an open top, the surface area $A$ consists of the area of the base and the four sides: $A = x^2 + 4 \cdot (x \cdot h)$

Substituting $h = \frac{97556}{x^2}$ into the surface area formula: $A(x) = x^2 + 4 \cdot \left( x \cdot \frac{97556}{x^2} \right)$ $A(x) = x^2 + \frac{390224}{x}$

Step 3: Differentiate to Find $A'(x)$

Now, find the derivative of $A(x)$ with respect to $x$: $A'(x) = 2x - \frac{390224}{x^2}$

Step 4: Set $A'(x) = 0$ to Find Critical Points

To minimize the surface area, set $A'(x) = 0$: $2x - \frac{390224}{x^2} = 0$

Multiply both sides by $x^2$ to clear the fraction: $2x^3 = 390224$ $x^3 = 195112$ $x = \sqrt[3]{195112}$ Calculating $x$, we get: $x \approx 58.4 \, \text{cm}$

Step 5: Use the Second Derivative Test

To confirm that this value of $x$ gives a minimum surface area, find the second derivative $A''(x)$: $A''(x) = 2 + \frac{780448}{x^3}$

Now, substitute $x \approx 58.4$ into $A''(x)$: $A''(58.4) = 2 + \frac{780448}{(58.4)^3}$ Since $A''(x) > 0$, the surface is concave up at this point, confirming a local minimum.

Final Answer:

• The side length $x \approx 58.4 \, \text{cm}$
• Height $h = \frac{97556}{x^2} \approx 28.6 \, \text{cm}$

Would you like further details on any of these steps?

Here are some related questions for deeper understanding:

1. How does changing the volume affect the dimensions that minimize surface area?
2. How would the problem differ if the box had a closed top?
3. What would happen to the surface area if the base shape changed from square to rectangular?
4. How can this method be applied to a cylinder with open top?
5. What if the material cost varied on the sides versus the base?

Tip: For optimization problems, always identify constraints and express variables in terms of one another to simplify.