An​ open-top cylindrical container is to have a volume

729

cmcubed

. 

What dimensions​ (radius and​ height) will minimize the surface​ area?

Question content area bottom

Part 1

The radius of the can is about

enter your response here

cm and its height is about

enter your response here
 
cm.

​(Do not round until the final answer. Then round to two decimal places as​ needed.)

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?

---

#### 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.

An open-top cylindrical container is to have a volume of 729 cm³. What dimensions (radius and height) will minimize the surface area? The radius of the can is about ___ cm and its height is about ___ cm. (Do not round until the final answer. Then round to two decimal places as needed.)

Explore how to find the dimensions (radius and height) that minimize the surface area of an open-top cylinder with a fixed volume of 729 cm³. This problem involves using calculus to derive formulas and optimize the cylinder's dimensions, making it suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation