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To solve this problem, we need to minimize the surface area of an open-top cylindrical can that can hold a fixed volume of 170 cm³. Let's go through the solution step-by-step.

### Step 1: Set up the Volume and Surface Area Formulas
1. **Volume of a cylinder**: \( V = \pi r^2 h \)
   - We know the volume \( V = 170 \) cm³, so:
     \[
     \pi r^2 h = 170
     \]
   
2. **Surface Area of a cylinder (with an open top)**: \( S = \pi r^2 + 2\pi rh \)
   - The first term \( \pi r^2 \) is the area of the circular base, and the second term \( 2\pi rh \) is the area of the side.

### Step 2: Express \( h \) in Terms of \( r \)
From the volume formula, solve for \( h \):
\[
h = \frac{170}{\pi r^2}
\]

### Step 3: Substitute \( h \) in the Surface Area Formula
Substitute \( h = \frac{170}{\pi r^2} \) into the surface area formula:
\[
S = \pi r^2 + 2\pi r \left(\frac{170}{\pi r^2}\right)
\]
Simplify this expression:
\[
S = \pi r^2 + \frac{340}{r}
\]

### Step 4: Minimize the Surface Area
To find the radius that minimizes the surface area, take the derivative of \( S \) with respect to \( r \), set it to zero, and solve for \( r \).

1. Differentiate \( S \) with respect to \( r \):
   \[
   \frac{dS}{dr} = 2\pi r - \frac{340}{r^2}
   \]
   
2. Set \( \frac{dS}{dr} = 0 \):
   \[
   2\pi r = \frac{340}{r^2}
   \]
   
3. Multiply both sides by \( r^2 \):
   \[
   2\pi r^3 = 340
   \]
   
4. Solve for \( r \):
   \[
   r^3 = \frac{340}{2\pi} = \frac{170}{\pi}
   \]
   \[
   r = \left(\frac{170}{\pi}\right)^{\frac{1}{3}}
   \]

### Step 5: Find the Corresponding Height \( h \)
Now, substitute \( r = \left(\frac{170}{\pi}\right)^{\frac{1}{3}} \) into the formula for \( h \):
\[
h = \frac{170}{\pi r^2}
\]
\[
h = \frac{170}{\pi \left(\left(\frac{170}{\pi}\right)^{\frac{1}{3}}\right)^2}
\]
\[
h = \frac{170}{\pi \cdot \left(\frac{170}{\pi}\right)^{\frac{2}{3}}}
\]
\[
h = \left(\frac{170}{\pi}\right)^{\frac{1}{3}}
\]

### Final Answer
- **Radius**: \( r = \left(\frac{170}{\pi}\right)^{\frac{1}{3}} \)
- **Height**: \( h = \left(\frac{170}{\pi}\right)^{\frac{1}{3}} \)

Both the radius and height are equal to \( \left(\frac{170}{\pi}\right)^{\frac{1}{3}} \).

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Would you like a further breakdown, or have any specific questions about this solution?

### Related Questions
1. What would change in the solution if the can had a closed top?
2. How does this approach change if the volume of the can were different, say 200 cm³?
3. Can we generalize this result for any volume \( V \)?
4. How do rational exponents simplify the calculation in problems like this?
5. How would the surface area formula look if the can were a cone instead of a cylinder?

### Tip
When dealing with optimization problems, always check if you can express one variable in terms of another to simplify the objective function.

A cylindrical can, open at the top, is to hold 170 cm³ of liquid. Find the height and radius that minimize the amount of material needed to manufacture the can. Enter your answer with rational exponents, and use π to represent pi.

Find the height and radius of an open-top cylindrical can that holds 170 cm³ of liquid, minimizing the material needed. This problem involves optimization, calculus, and geometry, with solutions using rational exponents and π.

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