a cylinder is to be designed to have a maximum combined length and circumference of 108 inches. use calculus to find the dimensions of the cylinder with maximum volume. then provide the maximum volume of the cylinder

To maximize the volume of a cylinder given a constraint on its combined length and circumference, we can proceed using calculus. Let's first define the variables:

- Let \( r \) be the radius of the base of the cylinder (in inches).
- Let \( h \) be the height of the cylinder (in inches).

### Step 1: Set up the constraint equation
We know that the combined length and circumference must be 108 inches:

\[
h + 2\pi r = 108
\]

### Step 2: Express height \( h \) in terms of \( r \)
From the constraint equation, we can solve for \( h \):

\[
h = 108 - 2\pi r
\]

### Step 3: Write the volume equation
The volume \( V \) of a cylinder is given by:

\[
V = \pi r^2 h
\]

Substitute \( h = 108 - 2\pi r \) into the volume equation:

\[
V(r) = \pi r^2 (108 - 2\pi r)
\]

Simplify:

\[
V(r) = \pi r^2 \cdot 108 - \pi r^2 \cdot 2\pi r
\]

\[
V(r) = 108\pi r^2 - 2\pi^2 r^3
\]

### Step 4: Maximize the volume using calculus
To find the value of \( r \) that maximizes the volume, we first take the derivative of \( V(r) \) with respect to \( r \) and set it equal to zero:

\[
\frac{dV}{dr} = \frac{d}{dr}\left(108\pi r^2 - 2\pi^2 r^3\right)
\]

\[
\frac{dV}{dr} = 216\pi r - 6\pi^2 r^2
\]

Set \( \frac{dV}{dr} = 0 \) to find the critical points:

\[
216\pi r - 6\pi^2 r^2 = 0
\]

Factor out \( r \):

\[
r(216\pi - 6\pi^2 r) = 0
\]

So, either \( r = 0 \) (which gives no volume) or:

\[
216\pi = 6\pi^2 r
\]

Solve for \( r \):

\[
r = \frac{216\pi}{6\pi^2} = \frac{216}{6\pi} = \frac{36}{\pi}
\]

Thus, the radius \( r = \frac{36}{\pi} \) inches.

### Step 5: Find the height \( h \)
Now that we have \( r = \frac{36}{\pi} \), we can substitute this value back into the equation for \( h \):

\[
h = 108 - 2\pi\left(\frac{36}{\pi}\right)
\]

Simplify:

\[
h = 108 - 72 = 36 \, \text{inches}
\]

### Step 6: Calculate the maximum volume
Now that we know \( r = \frac{36}{\pi} \) and \( h = 36 \), we can substitute these values into the volume formula:

\[
V = \pi r^2 h = \pi\left(\frac{36}{\pi}\right)^2 \cdot 36
\]

Simplify:

\[
V = \pi \cdot \frac{1296}{\pi^2} \cdot 36 = \frac{1296}{\pi} \cdot 36 = \frac{46656}{\pi} \, \text{cubic inches}
\]

### Final Answer
The dimensions of the cylinder that maximize the volume are:
- Radius \( r = \frac{36}{\pi} \approx 11.46 \) inches
- Height \( h = 36 \) inches

The maximum volume of the cylinder is approximately:

\[
V \approx \frac{46656}{\pi} \approx 14848.15 \, \text{cubic inches}
\]

Would you like more details on any part of the solution, or have any questions?

### Related Questions:
1. How does the constraint impact the cylinder's dimensions?
2. What happens to the volume if the height and radius are equal?
3. Can the same approach be used for minimizing surface area?
4. How does the volume change if the length-circumference sum is altered?
5. Can the problem be solved using a different method, like Lagrange multipliers?

### Tip:
In optimization problems, always define a clear constraint and objective function before taking derivatives. This helps in setting up the problem efficiently.

A cylinder is to be designed to have a maximum combined length and circumference of 108 inches. Use calculus to find the dimensions of the cylinder with maximum volume. Then provide the maximum volume of the cylinder.

Learn how to use calculus to optimize the dimensions of a cylinder with a maximum combined length and circumference of 108 inches. This problem covers key concepts in calculus and cylindrical geometry. Suitable for advanced high school or early college students.

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