A cylindrical can has a volume of 686 pi cm cubed. What dimensions yield the minimum surface​ area? 
Question content area bottom
Part 1
The radius of the can with the minimum surface area is 
  
enter your response here cm.
​(Simplify your​ answer.)
Part 2
The height of the can with the minimum surface area is 
  
enter your response here cm.
​(Simplify your​ answer.)

To solve for the dimensions that yield the minimum surface area of the cylindrical can, we use the following steps:

### Part 1: The Radius of the Can with Minimum Surface Area

Given the volume \( V = 686 \pi \, \text{cm}^3 \), the volume of a cylinder is calculated by:
\[
V = \pi r^2 h
\]
where \( r \) is the radius and \( h \) is the height.

From this equation, we can solve for \( h \) in terms of \( r \):
\[
h = \frac{V}{\pi r^2} = \frac{686 \pi}{\pi r^2} = \frac{686}{r^2}
\]

Next, the surface area \( S \) of a cylinder (including the top and bottom) is given by:
\[
S = 2\pi r^2 + 2\pi r h
\]
Substitute \( h = \frac{686}{r^2} \) into the surface area formula:
\[
S = 2\pi r^2 + 2\pi r \cdot \frac{686}{r^2}
\]
Simplify the expression:
\[
S = 2\pi r^2 + \frac{1372\pi}{r}
\]

To minimize the surface area, we need to take the derivative of \( S \) with respect to \( r \) and set it equal to zero:
\[
\frac{dS}{dr} = 4\pi r - \frac{1372\pi}{r^2}
\]
Set this equal to zero:
\[
4\pi r - \frac{1372\pi}{r^2} = 0
\]
Divide both sides by \( \pi \):
\[
4r - \frac{1372}{r^2} = 0
\]
Multiply both sides by \( r^2 \) to eliminate the fraction:
\[
4r^3 = 1372
\]
Solve for \( r \):
\[
r^3 = \frac{1372}{4} = 343
\]
\[
r = \sqrt[3]{343} = 7 \, \text{cm}
\]

### Part 2: The Height of the Can with Minimum Surface Area

Now that we have the radius, \( r = 7 \, \text{cm} \), substitute it into the equation for height:
\[
h = \frac{686}{r^2} = \frac{686}{7^2} = \frac{686}{49} = 14 \, \text{cm}
\]

### Final Answer:
- **Radius**: \( 7 \, \text{cm} \)
- **Height**: \( 14 \, \text{cm} \)

Would you like more details on any part of this process?

Here are 5 questions to expand on this:
1. What would happen to the surface area if the volume were doubled?
2. Can you derive the formula for the surface area of a cylinder from basic principles?
3. What is the significance of the derivative in finding the minimum surface area?
4. How does the ratio of height to radius affect the surface area?
5. Can this optimization technique be applied to other shapes?

**Tip:** Optimization problems often involve taking derivatives to find minimum or maximum values by identifying critical points.

A cylindrical can has a volume of 686 pi cm cubed. What dimensions yield the minimum surface area? Part 1: The radius of the can with the minimum surface area is _____ cm. Part 2: The height of the can with the minimum surface area is _____ cm.

Discover how to find the radius and height of a cylindrical can with a volume of 686 pi cubic cm that minimizes the surface area. This solution involves key geometric concepts and optimization techniques using calculus. Suitable for students in Grades 10-12.

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