To solve this problem, let's go through it step-by-step. The goal is to determine the dimensions of an open-top metal container with a square base that minimizes the amount of metal used. The volume of the container must be 32 cubic meters.

Problem Breakdown and Variables

1. Volume Constraint: The volume $V$ of the container is given by: $V = x^2 h = 32$ where $x$ is the side length of the square base and $h$ is the height of the container.

2. Surface Area (Metal Usage): Since the container is open at the top, the surface area $S$ that needs to be minimized (i.e., the amount of metal used) includes only the base and the four sides: $S = x^2 + 4xh$

Step 1: Express $h$ in Terms of $x$ Using the Volume Constraint

From the volume constraint $x^2 h = 32$, we can solve for $h$: $h = \frac{32}{x^2}$

Step 2: Substitute $h$ in the Surface Area Equation

Now, substitute $h = \frac{32}{x^2}$ into the surface area formula: $S = x^2 + 4x \left( \frac{32}{x^2} \right)$ Simplify this expression: $S = x^2 + \frac{128}{x}$

Step 3: Minimize the Surface Area Function

To minimize $S$, take the derivative of $S$ with respect to $x$ and set it equal to zero: $\frac{dS}{dx} = 2x - \frac{128}{x^2}$ Set the derivative equal to zero to find the critical points: $2x - \frac{128}{x^2} = 0$ Multiply through by $x^2$ to clear the denominator: $2x^3 = 128$ Solve for $x$: $x^3 = 64$ $x = \sqrt[3]{64} = 4$

Step 4: Find $h$ Using $x = 4$

Substitute $x = 4$ back into the equation for $h$: $h = \frac{32}{x^2} = \frac{32}{4^2} = \frac{32}{16} = 2$

Solution

The dimensions that minimize the amount of metal are:

• Base side length $x = 4$ meters
• Height $h = 2$ meters

Summary

The optimal dimensions for the container are a square base of 4 meters by 4 meters and a height of 2 meters.

Would you like further details on any of these steps or additional explanations?

1. What is the concept of minimizing a function, and how does it apply here?
2. How is the derivative used to find the minimum in optimization problems?
3. Why does having an open-top container affect the surface area calculation?
4. How would the problem change if the volume were different, such as 64 cubic meters?
5. What would happen to the surface area equation if the container had a lid?

Tip: In optimization problems, always start by identifying constraints and expressing one variable in terms of others to reduce the complexity of the function you are optimizing.