Problem 13 - Calculus: Early Transcendentals

Question

You want to make cylindrical containers to hold I liter using the least amount of construction material. The side is made from a rectangular piece of material, and this can be done with no material wasted. However, the top and bottom are cut from squares of side \(2 r,\) so that \(2(2 r)^{2}=8 r^{2}\) of material is needed ( rather than \(2 \pi r^{2},\) which is the total area of the top and bottom). Find the dimensions of the container using the least amount of material, and also find the ratio of height to radius for this container.

Step-by-Step Solution

Answer

The container dimensions are \( r = \frac{1}{2} \), \( h = \frac{4}{\pi} \), with a height-to-radius ratio of \( \frac{8}{\pi} \).

Step 1: Understand the problem

We are tasked to design a cylindrical container with a volume of 1 liter using the least amount of construction material. We must account for the fact that the tops and bottoms are cut from squares, not circles, so our total material calculation involves these squares.

Step 2: Set up the volume equation

The volume of the cylinder is given by the formula \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. Given that the volume must be 1 liter, we have \( \pi r^2 h = 1 \).

Step 3: Set up the material area equation

The material used consists of the sides (\( 2\pi rh \)) and the top and bottom squares (\( 8r^2 \)). Therefore, the total material area \( A \) is given by: \[ A = 2\pi rh + 8r^2. \]

Step 4: Substitute the expression for height

From \( \pi r^2 h = 1 \), solve for \( h \): \( h = \frac{1}{\pi r^2} \). Substitute \( h \) in the material area formula: \[ A = 2\pi r \frac{1}{\pi r^2} + 8r^2 = \frac{2}{r} + 8r^2. \]

Step 5: Minimize the material area

To find the dimensions that minimize the total material, we need to minimize \( A = \frac{2}{r} + 8r^2 \). Take the derivative with respect to \( r \) and set it to zero: \[ \frac{dA}{dr} = -\frac{2}{r^2} + 16r. \]

Step 6: Solve for critical points

Set the derivative equal to zero and solve for \( r \): \[ -\frac{2}{r^2} + 16r = 0 \] \[ 16r = \frac{2}{r^2} \] \[ 16r^3 = 2 \] \[ r^3 = \frac{1}{8} \] \[ r = \frac{1}{2} \].

Step 7: Solve for height

Using \( r = \frac{1}{2} \), substitute back to find \( h \): \[ h = \frac{1}{\pi r^2} = \frac{1}{\pi \left(\frac{1}{2}\right)^2} = \frac{4}{\pi} \].

Step 8: Find the height to radius ratio

The height to radius ratio is \( \frac{h}{r} \). Using \( h = \frac{4}{\pi} \) and \( r = \frac{1}{2} \): \[ \frac{h}{r} = \frac{\frac{4}{\pi}}{\frac{1}{2}} = \frac{8}{\pi} \].

Key Concepts

Cylinder Volume

When calculating the volume of a cylinder, the key elements include the radius of the circular base and the height of the cylinder. The formula for the volume is expressed as \( V = \pi r^2 h \). This formula comes from multiplying the area of the base, \( \pi r^2 \), with the height, \( h \). To solve any problems related to cylinder volumes, it's essential to have consistent units. For the given exercise, the volume is specified as 1 liter.

This means you have the equation \( \pi r^2 h = 1 \). Solving such equations requires manipulating the formula to express one variable in terms of another, which will assist when minimizing surface area later on. Understanding what each symbol represents and the relationship between the different parts of the cylinder's geometry is crucial for solving these types of problems effectively.

Surface Area Minimization

In the task at hand, we're trying to minimize the total material used for constructing a cylindrical container. If the tops and bottoms are circles, this would be straightforward, but here, they're cut from squares, which increases material use.

The equation for the total material needed becomes \( A = 2\pi rh + 8r^2 \). To minimize this surface area, we should understand that minimizing functions in calculus often involves finding the right dimensions or values of variables that yield the smallest possible result.

• Express height in terms of radius using the volume constraint \( \pi r^2 h = 1 \), resulting in \( h = \frac{1}{\pi r^2} \).
• Substitute \( h \) in the area equation to get \( A = \frac{2}{r} + 8r^2 \).

This rearrangement allows us to find the dimensions that minimize surface area using derivative techniques, which are primarily used to find points where the function is at a minimum or maximum.

Derivative Calculation

Derivatives help us find moments where a function reaches maximum or minimum values. In this situation, we're finding where the surface area function \( A = \frac{2}{r} + 8r^2 \) has a minimum.

By taking the derivative of \( A \) with respect to \( r \), we get:

• First, compute the derivative: \( \frac{dA}{dr} = -\frac{2}{r^2} + 16r \).
• To find critical points, set the derivative equal to zero, resulting in the equation \( -\frac{2}{r^2} + 16r = 0 \).

Solving that equation provides the values where the surface area could be minimized. Derivatives simplify the problem of quickly evaluating where curves increase, decrease, or stay constant, providing critical insights into optimization issues.

Critical Points

In optimization problems, critical points are where the derivative equals zero or is undefined. These points are candidates for local maxima, minima, or saddle points. For the surface area function \( A = \frac{2}{r} + 8r^2 \), we found its derivative as \( \frac{dA}{dr} = -\frac{2}{r^2} + 16r \).

Setting \( \frac{dA}{dr} = 0 \), we solve for \( r \):

• The equation simplifies to \( 16r = \frac{2}{r^2} \), then \( 16r^3 = 2 \).
• This results in \( r^3 = \frac{1}{8} \) and therefore \( r = \frac{1}{2} \).

Once \( r \) is known, using it in other formulas lets us find other dimensions, including \( h \), and establish the height-to-radius ratio. Critical points in calculus are powerful tools for pinpointing the exact values needed to meet optimization goals, like minimizing the surface area here.