- Radius: \(r \approx 4.04\) cm

- Height: \(h \approx 10.73\) cm

- Minimum cost: \(\approx 12.27\) cents

UpStudy Free Solution:

Express the height \(h\) as a function of the radius \(r\)

Given the volume of the cylinder:

\[V = \pi r^ 2 h = 550\]

Solving for \(h\):

\[h = \frac { 550} { \pi r^ 2} \]

Step 2: Express the cost as a function of the radius \(r\)

The material costs are as follows:

- The surface area of the sides: \(A_ { \text { sides} } = 2\pi rh\)

- The area of the bottom: \(A_ { \text { bottom} } = \pi r^ 2\)

- The area of the top: \(A_ { \text { top} } = \pi r^ 2\)

The cost function \(C\) is given by:

\[C = 0.03 \cdot 2\pi rh + 0.03 \cdot \pi r^ 2 + 0.05 \cdot \pi r^ 2\]

Substitute \(h = \frac { 550} { \pi r^ 2} \) into the cost function:

\[C = 0.03 \cdot 2\pi r \left ( \frac { 550} { \pi r^ 2} \right ) + ( 0.03 + 0.05) \cdot \pi r^ 2\]

\[C = 0.03 \cdot 2 \cdot 550 \cdot \frac { 1} { r} + 0.08 \cdot \pi r^ 2\]

\[C = 33 \cdot \frac { 1} { r} + 0.08 \pi r^ 2\]

Step 3: Take the derivative and find the critical points

To minimize the cost, we need to take the derivative of \(C\) with respect to \(r\) and set it to zero:

\[\frac { dC} { dr} = - 33 \cdot \frac { 1} { r^ 2} + 0.16\pi r\]

Set the derivative to zero:

\[- 33 \cdot \frac { 1} { r^ 2} + 0.16\pi r = 0\]

\[- 33 + 0.16\pi r^ 3 = 0\]

\[0.16\pi r^ 3 = 33\]

Solving for \(r\):

\[r^ 3 = \frac { 33} { 0.16\pi } \]

\[r^ 3 = \frac { 33} { 0.16 \cdot 3.14159} \]

\[r^ 3 \approx \frac { 33} { 0.502654} \]

\[r^ 3 \approx 65.65\]

\[r \approx \sqrt [ 3] { 65.65} \]

Using a calculator:

\[r \approx 4.04\]

Step 4: Calculate the height

Using \(r \approx 4.04\):

\[h = \frac { 550} { \pi ( 4.04) ^ 2} \]

\[h \approx \frac { 550} { \pi \cdot 16.32} \]

\[h \approx \frac { 550} { 51.29} \]

\[h \approx 10.73\]

Step 5: Calculate the minimum cost

Using \(r \approx 4.04\) and \(h \approx 10.73\) in the cost function:

\[C = 33 \cdot \frac { 1} { 4.04} + 0.08\pi ( 4.04) ^ 2\]

\[C \approx 33 \cdot 0.2475 + 0.08 \cdot 51.29\]

\[C \approx 8.17 + 4.10\]

\[C \approx 12.27 \text { cents} \]

Supplemental Knowledge

Understanding the principles of calculus and optimization is essential for solving problems involving the minimization or maximization of functions. Here are some key concepts and methods related to this topic:

Key Concepts:

1. Volume of a Cylinder:

- The volume \(V\) of a cylinder is given by:

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

- Here, \(r\) is the radius, and \(h\) is the height.

2. Surface Area of a Cylinder:

- The surface area \(A\) consists of:

- The lateral surface area (sides): \(A_ { \text { sides} } = 2\pi rh\)

- The area of the bottom: \(A_ { \text { bottom} } = \pi r^ 2\)

- The area of the top: \(A_ { \text { top} } = \pi r^ 2\)

3. Optimization in Calculus:

- Optimization involves finding the maximum or minimum value of a function within a given domain.

- Critical points are found where the first derivative \(f' ( x) \) is zero or undefined.

- The second derivative test can help determine whether these critical points are maxima, minima, or points of inflection.