To solve this problem, let’s break it down into steps, aiming to minimize the surface area of an open-top cylindrical container given a fixed volume.

Given Information and Variables

1. Volume constraint: $V = 729 \, \text{cm}^3$
   • Volume of a cylinder is given by: $V = \pi r^2 h$
2. Objective: Minimize the surface area of the open-top cylinder.
   • Surface area for an open-top cylinder: $\text{Surface Area} = \pi r^2 + 2\pi r h$

Here:

• $r$ is the radius of the base.
• $h$ is the height of the cylinder.

Step 1: Express Height in Terms of Radius

From the volume formula: $\pi r^2 h = 729$ Solve for $h$: $h = \frac{729}{\pi r^2}$

Step 2: Substitute $h$ into the Surface Area Formula

Now substitute $h$ from above into the surface area equation: $\text{Surface Area} = \pi r^2 + 2\pi r \left(\frac{729}{\pi r^2}\right)$ Simplify this: $\text{Surface Area} = \pi r^2 + \frac{1458}{r}$

Step 3: Minimize the Surface Area Function

To find the value of $r$ that minimizes the surface area, we take the derivative of the surface area with respect to $r$ and set it equal to zero.

1. Differentiate the surface area function with respect to $r$: $\frac{d}{dr} \left(\pi r^2 + \frac{1458}{r}\right) = 2\pi r - \frac{1458}{r^2}$

2. Set the derivative equal to zero to find critical points: $2\pi r - \frac{1458}{r^2} = 0$

3. Solve for $r$: $2\pi r^3 = 1458$ $r^3 = \frac{1458}{2\pi}$ $r = \sqrt[3]{\frac{1458}{2\pi}}$

4. Calculate $r$: $r \approx 6.2035 \, \text{cm}$

Step 4: Find $h$ Using the Value of $r$

Now, substitute $r \approx 6.2035$ cm back into the height formula: $h = \frac{729}{\pi (6.2035)^2}$ $h \approx 6.2035 \, \text{cm}$

Final Answer (rounded to two decimal places):

• Radius $r \approx 6.20$ cm
• Height $h \approx 6.20$ cm

Would you like further details on any of these steps?

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Related Questions:

1. How would the dimensions change if the volume increased to 1000 cm³?
2. How do we account for thickness in minimizing the surface area?
3. What happens to the dimensions if the cylinder has a closed top?
4. How would this problem change if surface area were maximized instead?
5. Could we minimize material cost by considering different materials for the base and side?

Tip:

When optimizing dimensions, remember to check if solutions make practical sense (e.g., height and radius values) for the physical context.