Problem 357 - Calculus Volume 3

Question

Find the maximum volume of a cylindrical soda can such that the sum of its height and circumference is 120 \(\mathrm{cm} .\)

Step-by-Step Solution

Answer

The maximum volume of the cylinder is \( \frac{64000}{\pi} \) cubic centimeters.

Step 1: Define the Variables and Formulae

Let's designate the radius of the cylinder as \( r \) and the height as \( h \). From the problem, we know the sum of the height \( h \) and the circumference, \( 2\pi r \), is 120 cm. Therefore, we have the equation: \[ h + 2\pi r = 120 \]. The volume of a cylinder is given by \( V = \pi r^2 h \).

Step 2: Express Height in Terms of Radius

From the equation \( h + 2\pi r = 120 \), we can express \( h \) in terms of \( r \): \[ h = 120 - 2\pi r \].

Step 3: Substitute for Height in the Volume Formula

Substitute \( h = 120 - 2\pi r \) from Step 2 into the volume formula \( V = \pi r^2 h \): \[ V = \pi r^2 (120 - 2\pi r) \]. Simplify this to: \[ V = 120\pi r^2 - 2\pi^2 r^3 \].

Step 4: Find the Derivative of the Volume Function

To find the radius that maximizes the volume, we need to take the derivative of \( V \) with respect to \( r \): \[ \frac{dV}{dr} = 240\pi r - 6\pi^2 r^2 \].

Step 5: Set Derivative Equal to Zero

Set the derivative \( \frac{dV}{dr} \) equal to zero and solve for \( r \): \[ 240\pi r - 6\pi^2 r^2 = 0 \]. Factor out \( 6\pi r \) to simplify: \[ 6\pi r (40 - \pi r) = 0 \]. This gives \( r = 0 \) or \( r = \frac{40}{\pi} \). Since \( r = 0 \) doesn't make sense for a real cylinder, we use \( r = \frac{40}{\pi} \).

Step 6: Determine the Height Using Radius

Using \( r = \frac{40}{\pi} \), substitute back into the equation for \( h \): \[ h = 120 - 2\pi \left(\frac{40}{\pi}\right) = 120 - 80 = 40 \].

Step 7: Calculate the Maximum Volume

Substitute \( r = \frac{40}{\pi} \) and \( h = 40 \) into the volume formula: \[ V = \pi \left(\frac{40}{\pi}\right)^2 \times 40 = \pi \times \frac{1600}{\pi^2} \times 40 = \frac{64000}{\pi} \].

Step 8: Verify Maximization

To ensure this is a maximum, check the second derivative \( \frac{d^2V}{dr^2} = 240\pi - 12\pi^2 r \). At \( r = \frac{40}{\pi} \), \( \frac{d^2V}{dr^2} < 0 \), confirming a maximum.

Key Concepts

Cylindrical Volume Maximization

To find the maximum volume of a cylinder, we need to focus on its geometric properties. The volume of a cylinder is calculated using the formula \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. When maximizing the volume under certain constraints, such as the sum of the height and circumference, it's necessary to express some variables in terms of others. In this exercise, the given constraint is \( h + 2\pi r = 120 \), which describes the relationship between the cylinder's height and its circumference. By manipulating this equation, you can substitute for one variable in the volume expression, allowing you to focus the optimization entirely on one variable, usually the radius \( r \). This streamlined focus leads to a more straightforward maximization process.

Derivative Applications

Derivatives are a powerful mathematical tool for finding maximum and minimum values, particularly in algebraic functions that describe real-world situations. When we seek to maximize the volume of a cylinder, we must find the derivative of the volume function with respect to the radius. This gives us \( \frac{dV}{dr} \), which indicates how changes in the radius affect the volume. By setting \( \frac{dV}{dr} = 0 \), we find critical points where the volume might achieve maximum or minimum values. Solving this equation reveals potential values of \( r \) where the volume could be optimized. This method is central to calculus as it provides insight into how variables interact to produce extreme values.

Constraint Optimization

In many real-world problems, such as the one involving the cylinder, we have given constraints that limit or affect the possibilities. Here, the constraint \( h + 2\pi r = 120 \) limits the relationships between the height and the circumference of the cylinder. Constraint optimization involves incorporating these restrictions into your calculations—guiding the substitution and simplification steps. This ensures that when calculating the maximum volume, you only consider solutions that satisfy the initial conditions of your problem. Successfully navigating constraints like this is crucial in making sure the optimization is both feasible and relevant to the problem at hand.

Second Derivative Test

The second derivative test is an essential tool for verifying whether a critical point found is a maximum or a minimum. After identifying the critical radius \( r = \frac{40}{\pi} \) in our problem, the second derivative \( \frac{d^2V}{dr^2} \) is calculated to confirm the nature of this critical point. For maximization, we want the second derivative to be negative at this point; \( \frac{d^2V}{dr^2} < 0 \), ensuring that we indeed have a maximum volume. This step is invaluable as it provides a mathematical assurance that our solution is not only mathematically correct but also practically optimal. The second derivative test is thus a reliable method to validate the conclusions drawn from our initial derivative calculations.