A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. The total volume of the solid is 6 cubic centimeters. Find the radius of the cylinder that produces the minimum surface area. (Round your answer to three decimal places.)


Let's solve the problem step by step.

### Step 1: Define the problem variables
- Let \( r \) be the radius of the cylinder (and the hemispheres).
- Let \( h \) be the height of the cylindrical part of the solid.

### Step 2: Express the total volume of the solid
The total volume of the solid consists of the volume of the cylinder plus the volume of the two hemispheres. Since two hemispheres make a whole sphere, the volume can be written as:

\[
V = \text{Volume of the cylinder} + \text{Volume of the sphere}
\]

\[
V = \pi r^2 h + \frac{4}{3} \pi r^3
\]

Given that the total volume \( V = 6 \) cubic centimeters, we have:

\[
\pi r^2 h + \frac{4}{3} \pi r^3 = 6
\]

### Step 3: Express the height \( h \) in terms of \( r \)

Solve the volume equation for \( h \):

\[
\pi r^2 h = 6 - \frac{4}{3} \pi r^3
\]

\[
h = \frac{6 - \frac{4}{3} \pi r^3}{\pi r^2}
\]

\[
h = \frac{6}{\pi r^2} - \frac{4r}{3}
\]

### Step 4: Express the surface area of the solid
The surface area \( S \) of the solid consists of the surface area of the cylindrical part and the surface area of the two hemispheres (which together form a sphere). The surface area can be written as:

\[
S = \text{Surface area of the cylinder} + \text{Surface area of the sphere}
\]

\[
S = 2\pi r h + 4\pi r^2
\]

Substitute the expression for \( h \):

\[
S = 2\pi r \left(\frac{6}{\pi r^2} - \frac{4r}{3}\right) + 4\pi r^2
\]

Simplify the expression:

\[
S = \frac{12}{r} - \frac{8\pi r^2}{3} + 4\pi r^2
\]

\[
S = \frac{12}{r} + \frac{4\pi r^2}{3}
\]

### Step 5: Find the critical points by taking the derivative
To minimize the surface area \( S \), take the derivative of \( S \) with respect to \( r \) and set it equal to zero:

\[
\frac{dS}{dr} = -\frac{12}{r^2} + \frac{8\pi r}{3}
\]

Set \(\frac{dS}{dr} = 0\):

\[
-\frac{12}{r^2} + \frac{8\pi r}{3} = 0
\]

Solve for \( r \):

\[
\frac{12}{r^2} = \frac{8\pi r}{3}
\]

\[
12 = \frac{8\pi r^3}{3}
\]

\[
r^3 = \frac{36}{8\pi}
\]

\[
r^3 = \frac{9}{2\pi}
\]

\[
r = \left(\frac{9}{2\pi}\right)^{\frac{1}{3}}
\]

### Step 6: Calculate the numerical value of \( r \)

\[
r = \left(\frac{9}{2\pi}\right)^{\frac{1}{3}} \approx \left(\frac{9}{6.283}\right)^{\frac{1}{3}} \approx \left(1.432\right)^{\frac{1}{3}} \approx 1.125
\]

Thus, the radius \( r \) that minimizes the surface area is approximately \( \boxed{1.125} \) cm.

Would you like more details or have any questions?

Here are some related questions you might find interesting:
1. How would the problem change if the total volume was different?
2. What is the minimum surface area in this case?
3. How does the height \( h \) vary with the radius for this minimum surface area?
4. Can we solve this problem using numerical optimization techniques?
5. What if the shape was not a cylinder with hemispheres but a different solid of revolution?

**Tip:** When solving optimization problems involving geometry, it often helps to express one variable in terms of others using known constraints, like volume or surface area.

A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. The total volume of the solid is 6 cubic centimeters. Find the radius of the cylinder that produces the minimum surface area. (Round your answer to three decimal places.)

Learn how to find the radius of a cylinder that minimizes the surface area when two hemispheres are joined to its ends, given a total volume of 6 cubic centimeters. This problem explores optimization in calculus and geometry, suitable for students in Grades 11-12.

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