Problem 345 - Calculus Volume 1

Question

Draw the given optimization problem and solve. Find the dimensions of the closed cylinder volume \(V=16 \pi\) that has the least amount of surface area.

Step-by-Step Solution

Answer

The cylinder has minimal surface area when \( r = 2 \) and \( h = 4 \).

Step 1: Understand the Problem

We have a closed cylinder with a volume constraint of \( V = 16\pi \). Our goal is to find the dimensions (radius \( r \) and height \( h \)) that minimize the surface area.

Step 2: Write Equations for Volume and Surface Area

The volume of a cylinder is given by \( V = \pi r^2 h \). The surface area of a closed cylinder is given by \( A = 2\pi r h + 2\pi r^2 \).

Step 3: Use the Volume Constraint

We know \( V = 16\pi \) which implies \( \pi r^2 h = 16\pi \). Simplifying gives \( r^2 h = 16 \), hence \( h = \frac{16}{r^2} \).

Step 4: Substitute \( h \) in Surface Area Formula

Substitute \( h = \frac{16}{r^2} \) into the surface area equation: \( A = 2\pi r \left(\frac{16}{r^2}\right) + 2\pi r^2 \). This simplifies to \( A = \frac{32\pi}{r} + 2\pi r^2 \).

Step 5: Differentiate the Surface Area with Respect to \( r \)

Differentiate \( A = \frac{32\pi}{r} + 2\pi r^2 \) with respect to \( r \): \( \frac{dA}{dr} = -\frac{32\pi}{r^2} + 4\pi r \).

Step 6: Find Critical Points

Set \( \frac{dA}{dr} = 0 \), giving \(-\frac{32\pi}{r^2} + 4\pi r = 0 \). Simplifying, we have \( 4\pi r^3 = 32\pi \). Thus, \( r^3 = 8 \), and \( r = 2 \).

Step 7: Find Corresponding Height

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

Step 8: Verify Minimum Value

To ensure this is a minimum, we check the second derivative \( \frac{d^2A}{dr^2} = \frac{64\pi}{r^3} + 4\pi \). For \( r = 2 \), the second derivative is positive, confirming a minimum.

Key Concepts

Cylinder Volume Formula

The volume of a cylinder is an essential concept in geometry and calculus. For a cylinder, the volume \( V \) is calculated using the formula:\[ V = \pi r^2 h \]Here:

• \( r \) is the radius of the cylinder's base.
• \( h \) is the height of the cylinder.

Calculating the volume involves multiplying the area of the circular base, which is \(\pi r^2\), by the height.
The formula provides a way to measure the space inside the cylinder. It's extremely valuable in real-world applications where cylindrical shapes are necessary, such as in engineering or manufacturing.
In our problem, knowing this formula allows us to work within the given constraint that the cylinder's volume is \(16\pi\). It ties directly into finding the optimal dimensions to satisfy a particular volume by allowing us to express one variable in terms of another.

Surface Area Minimization

Minimizing the surface area of a cylinder with a fixed volume can be particularly useful in manufacturing to reduce the material used, and ultimately, the cost. The surface area \( A \) for a closed cylinder is \[ A = 2\pi r h + 2\pi r^2 \]where:

• \( 2\pi r h \) represents the area of the side surface (also known as the lateral surface area).
• \( 2\pi r^2 \) accounts for the top and bottom circular surfaces.

To minimize the surface area, we substitute the expression for \( h \) recovered from the volume equation into the surface area formula. This provides a function \( A(r) = \frac{32\pi}{r} + 2\pi r^2 \), which depends only on the radius \( r \). By focusing on surface area optimization, we can solve for the dimensions that use the least amount of material possible.

Critical Points in Calculus

Critical points are where a function's derivative is zero or undefined. They are potential minimum or maximum points in calculus. To find these for the surface area, we differentiate the area function \( A(r) = \frac{32\pi}{r} + 2\pi r^2 \) with respect to \( r \):\[ \frac{dA}{dr} = -\frac{32\pi}{r^2} + 4\pi r \]Setting this derivative to zero, \(-\frac{32\pi}{r^2} + 4\pi r = 0\),helps us identify the critical points. Solving this equation gives \( r = 2 \), indicating critical points at this radius.
Identifying critical points is important because they inform us about the structure of the function. In optimization problems, such as minimizing costs or materials, critical points hold the key to finding the most efficient solution.

Second Derivative Test

The second derivative test is applied to confirm if a critical point is a minimum or maximum. It involves taking the second derivative of the function:\[ \frac{d^2A}{dr^2} = \frac{64\pi}{r^3} + 4\pi \]In our solution, we evaluate this at the known critical point \( r = 2 \). The outcome is \( \frac{d^2A}{dr^2} > 0 \), indicating the function is concave up at this point.

• If \( \frac{d^2A}{dr^2} > 0 \), the function has a local minimum.
• If \( \frac{d^2A}{dr^2} < 0 \), the function has a local maximum.

Hence, as applied here, the positive second derivative confirms that at \( r = 2 \), we achieve a minimal surface area. This is an essential step in verifying that the calculated dimensions actually provide the optimal solution, rather than just a potential one.