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To solve this problem, we need to determine the radius \( r \) and height \( h \) of a cylindrical can with a volume of 650 cm³, that minimizes the surface area of the can (since this will minimize the material needed). The can is open at the top, meaning the surface area consists only of the bottom circle and the cylindrical side.

### Step 1: Define the Volume of the Cylinder
The volume \( V \) of a cylinder is given by:
\[
V = \pi r^2 h
\]
We're told the volume is 650 cm³, so:
\[
\pi r^2 h = 650
\]
This can be rearranged to express \( h \) in terms of \( r \):
\[
h = \frac{650}{\pi r^2}
\]

### Step 2: Define the Surface Area to be Minimized
The surface area \( A \) of an open-top cylinder is the sum of the area of the bottom and the lateral surface area:
\[
A = \pi r^2 + 2 \pi r h
\]
Substitute \( h \) from the volume equation:
\[
A = \pi r^2 + 2 \pi r \left( \frac{650}{\pi r^2} \right)
\]
Simplify the equation:
\[
A = \pi r^2 + \frac{1300}{r}
\]

### Step 3: Find the Minimum Surface Area
To find the value of \( r \) that minimizes \( A \), take the derivative of \( A \) with respect to \( r \) and set it to zero:
\[
\frac{dA}{dr} = 2\pi r - \frac{1300}{r^2}
\]
Set \(\frac{dA}{dr} = 0\):
\[
2\pi r = \frac{1300}{r^2}
\]
Multiply both sides by \( r^2 \):
\[
2\pi r^3 = 1300
\]
Solve for \( r \):
\[
r^3 = \frac{1300}{2\pi} = \frac{650}{\pi}
\]
\[
r = \left( \frac{650}{\pi} \right)^{\frac{1}{3}}
\]

### Step 4: Calculate \( h \)
Substitute \( r = \left( \frac{650}{\pi} \right)^{\frac{1}{3}} \) back into the equation for \( h \):
\[
h = \frac{650}{\pi r^2} = \frac{650}{\pi \left( \frac{650}{\pi} \right)^{\frac{2}{3}}}
\]
Simplify:
\[
h = \left( \frac{650}{\pi} \right)^{\frac{1}{3}}
\]

### Final Answer
So the values for \( r \) and \( h \) that minimize the surface area are:
\[
\text{Radius } r = \left( \frac{650}{\pi} \right)^{\frac{1}{3}}
\]
\[
\text{Height } h = \left( \frac{650}{\pi} \right)^{\frac{1}{3}}
\]

Would you like further explanation on any part of this solution?

### Follow-up Questions
1. How would the formula change if the can were closed on top?
2. What would the effect be on \( r \) and \( h \) if the volume were doubled?
3. How can we verify if this solution gives a minimum or maximum surface area?
4. What is the approximate decimal value of \( \left( \frac{650}{\pi} \right)^{\frac{1}{3}} \)?
5. How does this approach apply to other shapes with fixed volumes?

### Tip
When working with optimization problems, always define your constraints first and express one variable in terms of the others to reduce complexity.

A cylindrical can, open at the top, is to hold 650 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 pi to represent π.

Learn how to find the radius and height that minimize the material needed for a cylindrical can, open at the top, with a fixed volume of 650 cm³. This solution uses calculus-based optimization and covers formulas for volume and surface area of a cylinder. Ideal for high school students in Grades 11-12.

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