Problem 17 - Calculus for Biology and Medicine

Question

Find the dimensions of a right circular cylindrical can (with bottom and top closed) that has a volume of 1 liter and that minimizes the amount of material used. (Note: One liter corresponds to \(1000 \mathrm{~cm}^{3}\).)

Step-by-Step Solution

Answer

Optimal dimensions are radius \( r \approx 5.42 \) cm and height \( h \approx 10.85 \) cm.

Step 1: Define the Problem

We are tasked with finding the dimensions (radius and height) of a cylindrical can that has a volume of 1 liter (1000 cm³) and minimizes surface area, thus using the least amount of material.

Step 1: Write the Volume Equation

The volume of a cylinder is given by the formula: \[ V = \pi r^2 h \] Where \( V = 1000 \) cm³ (1 liter). Substituting this, we get: \[ 1000 = \pi r^2 h \]

Step 2: Express Height in Terms of Radius

From the volume equation \( 1000 = \pi r^2 h \), solve for \( h \):\[ h = \frac{1000}{\pi r^2} \]

Step 3: Write the Surface Area Equation

The surface area \( A \) of a cylinder (closed on top and bottom) is given by: \[ A = 2\pi r^2 + 2\pi rh \] The first term is the area of the two circular ends, and the second term is the lateral surface area.

Step 4: Substitute h in Surface Area Equation

Substitute the expression for \( h \) from Step 2 into the surface area equation:\[ A = 2\pi r^2 + 2\pi r \left( \frac{1000}{\pi r^2} \right) \] Simplify this to get:\[ A = 2\pi r^2 + \frac{2000}{r} \]

Step 5: Find the Derivative of Surface Area

Differentiate the surface area function with respect to \( r \): \[ \frac{dA}{dr} = 4\pi r - \frac{2000}{r^2} \]

Step 6: Set Derivative to Zero for Optimization

To minimize \( A \), set \( \frac{dA}{dr} = 0 \) and solve for \( r \): \[ 4\pi r = \frac{2000}{r^2} \] Multiply both sides by \( r^2 \) and simplify: \[ 4\pi r^3 = 2000 \] \[ r^3 = \frac{2000}{4\pi} \] \[ r^3 = \frac{500}{\pi} \] \[ r = \left( \frac{500}{\pi} \right)^{1/3} \]

Step 7: Calculate Optimal Radius

Compute \( r \) using the value derived:\[ r \approx \left( \frac{500}{\pi} \right)^{1/3} \] This approximates to \( r \approx 5.42 \) cm.

Step 8: Calculate Corresponding Height

Using the volume equation, calculate \( h \): \[ h = \frac{1000}{\pi (5.42)^2} \] \[ h \approx 10.85 \] cm.

Step 9: Verify Optimization

Double-check the calculations to ensure that calculated \( r \) and \( h \) provide the required volume and minimize the surface area.The calculation shows that this does indeed minimize the surface area.

Key Concepts

Cylinder Volume Formula

In calculus, understanding how to calculate the volume of shapes like a cylinder is essential, especially in optimization problems. The volume of a cylindrical object is determined by the formula: \[ V = \pi r^2 h \]where:

• \( V \) is the volume,
• \( r \) is the radius of the base,
• \( h \) is the height.

To find the volume, multiply the area of the circular base \( \pi r^2 \) by the height \( h \). This basic but powerful formula is the starting point for finding dimensions when given a specific volume, like 1 liter, which equates to 1000 cm³ for a cylinder.
In an exercise where you're tasked with finding the dimensions of a cylinder that fulfills a volume requirement, you begin by defining this relationship. Solving for one variable in terms of another can help simplify the problem, especially when involving complex equations like those in optimization tasks.

Surface Area Minimization

Minimizing surface area is a common objective in various applications, such as designing containers with less material cost. For a closed cylindrical can, the surface area \( A \) includes two components:

• The area of the two circular ends: \( 2\pi r^2 \)
• The lateral surface area: \( 2\pi rh \)

The formula for surface area is:\[ A = 2\pi r^2 + 2\pi rh \].
To minimize this surface area while maintaining a set volume, it's necessary to express all surface area terms in terms of a single variable. This often involves substituting expressions of \( h \) or \( r \) derived from the volume equation. Here, representing height \( h \) as \( \frac{1000}{\pi r^2} \) allows us to only focus on \( r \) in the surface area equation:\[ A = 2\pi r^2 + \frac{2000}{r} \].
By differentiating this expression with respect to \( r \) and equating to zero, we find the optimal radius that minimizes the surface area. Solving for this radius determines the dimensions that require the least amount of material, a valuable outcome in manufacturing and design.

Applications of Derivatives

Derivatives are crucial in optimization problems, particularly when determining maxima or minima of functions. In the context of minimizing surface area for a cylinder, derivatives help identify the point where the slope of the surface area function is zero, indicating a minimum or maximum point.
The derivative of the surface area function \( A \) with respect to \( r \) in this case is:\[ \frac{dA}{dr} = 4\pi r - \frac{2000}{r^2} \].
Setting this derivative to zero:\[ 4\pi r - \frac{2000}{r^2} = 0 \]helps us find critical points which are further analyzed for minimum conditions. Solving this equation gives the radius value that minimizes the surface area while adhering to volume constraints.
The use of derivatives in this manner underlies broader applications across many scientific and engineering fields, including economics, where cost or resource optimization is paramount. Derivatives provide the tools needed to explore relationships between variables and ensure efficient and effective solutions.