To solve this problem, let's break it down step-by-step.

Problem Breakdown

We are given:

1. A box with a square base and an open top.
2. The volume of the box is $119164 \, \text{cm}^3$.
3. We need to find the dimensions of the box (specifically $x$, the side length of the square base) that minimizes the amount of material used.

Step 1: Set up the variables and relationships

1. Let $x$ be the length of one side of the square base.
2. Let $y$ be the height of the box.

Since the volume of the box is given by: $x^2 \cdot y = 119164$ we can solve for $y$ in terms of $x$: $y = \frac{119164}{x^2}$

Step 2: Set up the surface area function

The surface area $A(x)$ of the box is the amount of material used. Since the box has an open top, it consists of:

1. The base area: $x^2$
2. The four side walls: $4 \cdot (x \cdot y)$

Substitute $y$ from the previous step: $A(x) = x^2 + 4 \cdot x \cdot \frac{119164}{x^2}$ Simplify this expression: $A(x) = x^2 + \frac{4 \cdot 119164}{x}$ $A(x) = x^2 + \frac{476656}{x}$

Step 3: Find the minimum of $A(x)$

To find the value of $x$ that minimizes $A(x)$, we need to take the derivative $A'(x)$ and set it to zero.

1. Differentiate $A(x)$: $A'(x) = 2x - \frac{476656}{x^2}$

2. Set $A'(x) = 0$ to find the critical points: $2x - \frac{476656}{x^2} = 0$ $2x^3 = 476656$ $x^3 = 238328$ $x = \sqrt[3]{238328}$

After calculating $x \approx 62$ cm.

Step 4: Find $y$

Using $x = 62$ cm in the volume equation to find $y$: $y = \frac{119164}{62^2}$ $y \approx 31 \, \text{cm}$

Step 5: Minimum Surface Area Calculation

Now, substitute $x = 62$ into $A(x)$ to get the minimum surface area: $A(62) = 62^2 + \frac{476656}{62}$ Calculating this gives the minimum surface area needed.

Would you like further details or clarification on any of these steps? Here are some related questions for practice:

1. How do you find the volume of a box with a different shaped base?
2. What changes in calculations if the box has a closed top?
3. How do derivatives help in finding minimum values in optimization problems?
4. What would be the surface area function if the volume or dimensions were different?
5. How do constraints affect optimization in different shapes?

Tip: In optimization problems, always isolate one variable using given constraints to simplify the function before differentiating.