Problem 18 - Calculus for Biology and Medicine

Question

Find the dimensions of a right circular cylinder that is open on the top, is closed on the bottom, holds 1 liter, and uses the least amount of material.

Step-by-Step Solution

Answer

The optimal dimensions are a radius of approximately 5.42 cm and a height of approximately 10.84 cm.

Step 1: Understand the Problem

We need to find the dimensions (radius and height) of a right circular cylinder that has no top and must hold a volume of 1 liter (1000 cm³). Moreover, we have to minimize the material used, which means finding the dimensions that minimize the surface area of the cylinder.

Step 2: Model the Cylinder

The surface area of the cylinder, including the base but not the top, is given by the formula: \[ A = \ pi r^2 + 2 \pi rh \]where \( r \) is the radius and \( h \) is the height of the cylinder. The volume of the cylinder is given by: \[ V = \pi r^2 h \]which equals 1000 cm³.

Step 3: Express Height in Terms of Radius

Using the volume formula, express \( h \) in terms of \( r \): \[ h = \frac{1000}{\pi r^2} \]Substitute this expression for \( h \) into the surface area formula to make \( A \) a function of \( r \) only.

Step 4: Surface Area Function of Radius

Substitute \( h = \frac{1000}{\pi r^2} \) into the surface area formula to get \[ A(r) = \pi r^2 + 2 \pi r \frac{1000}{\pi r^2} \]Simplify the expression: \[ A(r) = \pi r^2 + \frac{2000}{r} \]

Step 5: Differentiate and Minimize

Find the derivative of \( A(r) \) with respect to \( r \), set it to zero, and solve for \( r \): \[ A'(r) = 2\pi r - 2000 \frac{1}{r^2} \]Set \( A'(r) = 0 \): \[ 2\pi r = \frac{2000}{r^2} \]Multiply both sides by \( r^2 \) and solve for \( r \).

Step 6: Solve for Optimal Radius

After clearing the equation: \[ 2\pi r^3 = 2000 \]we find that \[ r^3 = \frac{1000}{\pi} \]Leading to \[ r = \sqrt[3]{\frac{1000}{2\pi}} \].

Step 7: Calculate Optimal Height

Use the optimal \( r \) to find \( h \) using the formula \[ h = \frac{1000}{\pi r^2} \].

Step 8: Verify Solution

Ensure that the solution satisfies both the constraint on volume and minimizes the surface area by checking the second derivative or using alternative methods.

Key Concepts

Surface Area Minimization

Minimizing the surface area of a shape often involves optimizing the material used for its construction. In our cylinder problem, this means using the least material possible to contain a specific volume, which is 1 liter or 1000 cm³.

When optimizing for surface area, especially for objects like cylinders, we aim to reduce the unnecessary or excess surface that could be costly or inefficient. The cylinder in question has a unique feature as it is open on top and closed at the bottom. This creates different constraints for calculating the area versus a closed cylinder.

The surface area for such a cylinder is given by the formula:

• The area of the base: \( \pi r^2 \).
• The side or lateral surface area: \( 2 \pi rh \).

We aim to find the dimensions that keep the volume consistent while reducing \( A = \ \pi r^2 + 2 \pi rh \) as much as possible.

Cylinder Volume Calculation

In geometry and calculus, determining the volume of a cylinder is a fundamental concept. For a right circular cylinder, the volume can be expressed in terms of the radius and height. This is calculated using the formula:

• \( V = \pi r^2 h \)

For the cylinder to hold a specific volume, like 1000 cm³ in our example, we have a constraint that directly influences the dimensions of the cylinder.

Given this constraint, we can express one variable in terms of another. Here, we did this by finding height () \( h \) as a function of radius () \( r \):

• \( h = \frac{1000}{\ \pi r^2} \)

This substitution is critical as it reduces the complexity of the optimization process. Having a single variable expression makes differentiating easier which is vital in finding the optimal dimensions that satisfy both surface area minimization and volume constraints.

Differentiation Techniques

Differentiation is a key calculus technique used to find rates of change and to optimize functions. In this problem, we differentiated the surface area function () \( A(r) = \pi r^2 + \frac{2000}{r} \) to find critical points.

Taking the derivative allows us to pinpoint where the surface area reaches its minimum. By setting the derivative equal to zero, we isolate the critical points. We differentiated the function as follows:

• \( A'(r) = 2\ \pi r - 2000 \frac{1}{r^2} \)

To solve for () \( r \), we set () \( A'(r) = 0 \), and solved the equation to find the optimal radius. Once we have a critical point for () \( r \), we continue by calculating the second derivative to confirm that the critical point indeed corresponds to a minimum, rather than a maximum or a saddle point.

This ensures our solution is correct and satisfies the problem's optimization criteria.