Problem 345 - Calculus Volume 1

Question

For the following exercises, draw the given optimization problem and solve. Find the dimensions of the closed cylinder volume \(V=16 \pi\) that has the least amount of sulface area.

Step-by-Step Solution

Answer

The dimensions are radius \( r = 2 \) and height \( h = 4 \).

Step 1: Define the problem

We are given a closed cylinder with a fixed volume of \( V = 16 \pi \) and asked to find the dimensions (radius \( r \) and height \( h \)) that minimize its surface area.

Step 2: Set up the equations

The volume of a cylinder \( V \) is given by the formula \( V = \pi r^2 h \). We know that \( V = 16 \pi \). The surface area \( S \) of a closed cylinder is given by \( S = 2\pi r^2 + 2\pi rh \).

Step 3: Express one variable in terms of the other

Since \( V = \pi r^2 h \), we can express \( h \) in terms of \( r \): \( h = \frac{16}{r^2} \). Substitute \( h \) into the surface area equation: \( S = 2\pi r^2 + 2\pi r \frac{16}{r^2} \).

Step 4: Simplify the surface area equation

The expression becomes \( S = 2\pi r^2 + \frac{32\pi}{r} \). Now, we have the surface area as a function of \( r \): \( S = 2\pi r^2 + \frac{32\pi}{r} \).

Step 5: Differentiate the surface area with respect to \( r \)

Find the derivative of \( S \) with respect to \( r \), \( \frac{dS}{dr} = 4\pi r - \frac{32\pi}{r^2} \). This will help find the critical points which can give the minimum surface area.

Step 6: Find the critical points

Set \( \frac{dS}{dr} = 0 \): \( 4\pi r - \frac{32\pi}{r^2} = 0 \). Solve for \( r \): \( 4\pi r^3 = 32\pi \), which simplifies to \( r^3 = 8 \) and hence \( r = 2 \).

Step 7: Determine the corresponding height

Using \( h = \frac{16}{r^2} \) with \( r = 2 \), we get \( h = \frac{16}{4} = 4 \).

Step 8: Verify if it is a minimum

To confirm that \( r = 2 \) gives a minimum, check the second derivative \( \frac{d^2S}{dr^2} = 4\pi + \frac{64\pi}{r^3} \). For \( r = 2 \), this is positive, indicating a local minimum.

Key Concepts

Cylinder Volume

A cylinder is a common geometric shape you might encounter in your mathematics or science classes. It consists of two circular bases and a curved surface connecting them. When considering the volume of a cylinder, you're essentially measuring the amount of space it encloses. The formula for the volume of a cylinder is crucial here:
\[ V = \pi r^2 h \]Where:

• \( V \) is the volume
• \( r \) is the radius of the circular base
• \( h \) is the height of the cylinder

In our problem, we're given a fixed volume, \( V = 16 \pi \). This means any cylinder we consider must have a volume calculated using this formula, leading to constraints on the possible values of \( r \) and \( h \). Understanding how to manipulate and use this formula is essential for further calculations, especially when integrating it with other conditions such as minimizing surface area.

Surface Area Minimization

Minimizing surface area involves finding the dimensions that result in the least possible outer surface. For a closed cylinder, the surface area, \( S \), includes both the sides and the ends:
\[ S = 2\pi r^2 + 2\pi rh \]This equation is the sum of the area of two circular bases and the rectangular section formed when you "unwrap" the curved surface of the cylinder. The problem asks to minimize this surface area given a fixed volume. By substituting the relationship \( h = \frac{16}{r^2} \) into the area formula, we target a function of \( r \) alone:
\[ S = 2\pi r^2 + \frac{32\pi}{r} \]Now, with this function, we can use calculus to find the optimal values of \( r \) and \( h \) to achieve the smallest surface area possible. This step is critical in manufacturing and architectural designs, where material costs are minimized.

Critical Points in Calculus

Critical points are vital in any optimization problem since they signal potential minimums or maximums in a function. They occur where the derivative of a function equals zero or is undefined. Our task is to find such points for the surface area function:
\[ \frac{dS}{dr} = 4\pi r - \frac{32\pi}{r^2} \]Setting \( \frac{dS}{dr} = 0 \) gives:\[ 4\pi r - \frac{32\pi}{r^2} = 0 \]This equation simplifies to find \( r = 2 \). This \( r \) gives us a critical point. For a complete analysis, you need to verify whether this point indeed gives a minimum by further investigation, which brings us to the second derivative test.

Second Derivative Test

The second derivative test helps confirm the nature of critical points. If the second derivative is positive at a critical point, the function has a local minimum there. For our surface area function, the second derivative is:
\[ \frac{d^2S}{dr^2} = 4\pi + \frac{64\pi}{r^3} \]By substituting \( r = 2 \), we get a positive value:\[ \frac{d^2S}{dr^2} \bigg|_{r=2} = 4\pi + \frac{64\pi}{8} > 0 \]Therefore, the critical point \( r = 2 \) is a local minimum for the surface area.
This confirms that the dimensions with \( r = 2 \) and \( h = 4 \) give the least surface area for the cylinder with a volume \( V = 16 \pi \). Understanding and applying the second derivative test is essential in confirming the nature of critical points, ensuring you have found an optimal solution.