Problem 12 - Calculus: Early Transcendentals

Question

For a cylinder with given surface area \(S,\) including the top and the bottom, find the ratio of height to base radius that maximizes the volume.

Step-by-Step Solution

Answer

The ratio of height to base radius is 2:1.

Step 1: Understand the Problem

We have a cylinder with a given surface area which includes both the top and bottom surfaces. We are tasked with finding the ratio of height to base radius that maximizes the volume of the cylinder.

Step 2: Express Surface Area Formula

The total surface area of a cylinder is given by the formula: \[ S = 2\pi r h + 2\pi r^2 \]where \( r \) is the base radius, and \( h \) is the height of the cylinder. This equation considers the curved surface and both circular ends.

Step 3: Express Volume Formula

The volume of a cylinder is given by the formula: \[ V = \pi r^2 h \]Our goal is to maximize \( V \) by finding the optimal ratio of \( h \) to \( r \), given a fixed surface area \( S \).

Step 4: Solve for Height in terms of Radius

Rearrange the surface area equation to express \( h \) in terms of \( r \):\[ 2\pi r h = S - 2\pi r^2 \]\[ h = \frac{S - 2\pi r^2}{2\pi r} \]

Step 5: Substitute for Height in Volume Formula

Substitute the expression for \( h \) from Step 4 into the volume formula: \[ V = \pi r^2 \left( \frac{S - 2\pi r^2}{2\pi r} \right) \]\[ V = \frac{r(S - 2\pi r^2)}{2} \]This represents the volume as a function of the radius \( r \).

Step 6: Differentiate and Find Critical Points

To maximize the volume, take the derivative of \( V \) with respect to \( r \), set it to zero, and solve for \( r \):\[ V'(r) = \frac{d}{dr} \left( \frac{r(S - 2\pi r^2)}{2} \right) = 0 \]This derivative will produce a critical point which indicates the value of \( r \) that maximizes \( V \).

Step 7: Simplify Expression

From Step 6, calculate the derivative: \[ V'(r) = \frac{1}{2} (S - 6\pi r^2) \]Set the derivative equal to zero:\[ S - 6\pi r^2 = 0 \]\[ r^2 = \frac{S}{6\pi} \]Thus, \( r = \sqrt{\frac{S}{6\pi}} \).

Step 8: Calculate Height and Ratio

Substitute \( r = \sqrt{\frac{S}{6\pi}} \) back into the equation for \( h \):\[ h = \frac{S - 2\pi r^2}{2\pi r} = \frac{S - 2\pi \left( \frac{S}{6\pi} \right)}{2\pi \left( \sqrt{\frac{S}{6\pi}}\right)} \]\[ h = \frac{S - \frac{S}{3}}{2\pi \sqrt{\frac{S}{6\pi}}} \]\[ h = \frac{2S}{3} \cdot \frac{1}{2\pi \sqrt{\frac{S}{6\pi}}} \]Simplify to find the ratio of height to radius: \[ \frac{h}{r} = 2 \]

Step 9: Conclusion

The ratio of height to base radius that maximizes the volume of a cylinder with a given surface area is 2:1. This conclusion follows from the simplified expression derived from earlier calculations.

Key Concepts

Surface Area of Cylinder

To optimize geometric shapes such as cylinders, it's crucial to understand how to express and use the formula for surface area. For a cylinder, the total surface area \( S \) includes the surface area wrapped around the body plus the surface areas of the top and bottom circles. This is given by the formula:

• The area of the rectangular side (curved surface) is \( 2\pi r h \).
• The area of the top and bottom (two circles) is \( 2\pi r^2 \).

Together, the surface area formula is \( S = 2\pi r h + 2\pi r^2 \). Understanding this formula allows us to see how changes in height \( h \) and radius \( r \) will affect the overall surface area of the cylinder, holding one element constant while changing the other.

Volume of Cylinder

The volume of a cylinder helps determine the amount of space contained within. It is represented by the formula \( V = \pi r^2 h \), where:

• \( \pi r^2 \) is the area of the circular base.
• \( h \) is the height of the cylinder.

This formula shows that the volume depends on both the square of the radius and the height. When optimizing for maximum volume, knowing how these two variables interact under the constraints of a fixed surface area is key. The formula becomes significantly vital in producing a maximum with respect to the ratio \( \frac{h}{r} \).

Differentiation for Optimization

Differentiation is a powerful tool used in calculus to find maximum and minimum values of functions, and it's crucial for solving optimization problems. By taking the derivative of the volume function and setting it to zero, we can locate critical points that suggest where these extremum points (highest or lowest) occur.For the volume of a cylinder as a function of radius \( r \), differentiation with respect to \( r \) gives:\[ V'(r) = \frac{1}{2} (S - 6\pi r^2) \]Setting \( V'(r) = 0 \) helps us solve for \( r \) and find where the volume peaks. This type of calculus application shows us how the radius relates to maximizing the volume within the given constraints. Calculation of the derivatives involves using rules to handle expressions with multiplication and powers, and it's an essential part of the optimization process to find the desired ratio.

Ratio of Height to Radius

The optimization challenge posed is finding the ratio of height \( h \) to base radius \( r \) that maximizes volume given a fixed surface area. From the differentiation process, after solving for the radius, we find:\[ r = \sqrt{\frac{S}{6\pi}} \]Substituting this back to find height gives:\[ h = \frac{2S}{3} \cdot \frac{1}{2\pi \sqrt{\frac{S}{6\pi}}} \]When simplified, this results in the ratio \( \frac{h}{r} = 2 \). This calculation shows the importance of how altering one parameter (while keeping the surface area constant) impacts maximizing another property, like volume, providing valuable insights into the symbiotic nature of geometric dimensions in optimization problems.