To solve this problem, let’s break it down into steps.

Problem Analysis

We have a box with:

1. A square base.
2. An open top.
3. A required volume of $2500 \ \text{cm}^3$.

We want to find the dimensions that minimize the surface area of this box.

Definitions and Variables

Let:

• $x$ be the length of a side of the square base (in cm).
• $h$ be the height of the box (in cm).

The volume $V$ of the box is given by: $V = x^2 \cdot h$

Since $V = 2500 \ \text{cm}^3$, we have: $x^2 \cdot h = 2500$ Solving for $h$: $h = \frac{2500}{x^2}$

Surface Area Calculation

Since the box has an open top, the surface area $S$ consists of the area of the base and the area of the four sides: $S = x^2 + 4 \cdot (x \cdot h)$ Substitute $h = \frac{2500}{x^2}$ into this equation: $S = x^2 + 4 \cdot \left(x \cdot \frac{2500}{x^2}\right)$ Simplify the equation: $S = x^2 + \frac{10000}{x}$

Minimizing the Surface Area

To find the minimum surface area, take the derivative of $S$ with respect to $x$ and set it to zero:

1. Differentiate $S$ with respect to $x$: $\frac{dS}{dx} = 2x - \frac{10000}{x^2}$
2. Set the derivative to zero: $2x - \frac{10000}{x^2} = 0$ Rearrange to solve for $x$: $2x^3 = 10000$ $x^3 = 5000$ $x = \sqrt[3]{5000} \approx 17.1$

Solving for $h$

Using $x \approx 17.1$ cm in $h = \frac{2500}{x^2}$: $h = \frac{2500}{(17.1)^2} \approx 8.55$

Calculating the Minimum Surface Area

Now, substitute $x \approx 17.1$ and $h \approx 8.55$ back into the surface area formula: $S = (17.1)^2 + \frac{10000}{17.1} \approx 292 + 585 \approx 877 \ \text{cm}^2$

Answer

The minimum possible surface area, rounded to the nearest whole number, is: $\boxed{877}$

Would you like a detailed breakdown of any step or further questions?

Further Related Questions

1. How do we determine if this value is a minimum using the second derivative test?
2. What would the surface area be if the box had a closed top?
3. How does changing the volume affect the minimum surface area?
4. Can we use calculus techniques to minimize other constraints, such as perimeter?
5. How does the cube root calculation affect the accuracy of the result?

Tip

To minimize problems involving volume and surface area, try expressing one variable in terms of the other using given constraints. Then, apply calculus to find extreme values.