Problem 88 - College Algebra

Question

For the following exercises, construct a rational function that will help solve the problem. Then, use a calculator to answer the question. A right circular cylinder is to have a volume of 40 cubic inches. It costs 4 cents/square inch to construct the top and bottom and 1 cent/square inch to construct the rest of the cylinder. Find the radius to yield minimum cost. Let \(x=\) radius.

Step-by-Step Solution

Answer

The radius that yields minimum cost is approximately 1.12 inches.

Step 1: Volume of a Cylinder

The volume \( V \) of a right circular cylinder can be expressed as \( V = \pi x^2 h \), where \( x \) is the radius and \( h \) is the height. Given \( V = 40 \), we have the equation \( \pi x^2 h = 40 \). Solving for \( h \) gives \( h = \frac{40}{\pi x^2} \).

Step 2: Surface Area of a Cylinder

The surface area \( A \) of a cylinder is composed of the top and bottom circles (\( 2\pi x^2 \)) and the side (\( 2\pi x h \)). Substituting \( h \) from Step 1: \[ A = 2\pi x^2 + 2\pi x \left( \frac{40}{\pi x^2} \right) = 2\pi x^2 + \frac{80}{x}. \]

Step 3: Cost Function

The cost \( C \) to construct the cylinder is calculated by \( C = 4 \times (\text{area of top and bottom}) + 1 \times (\text{area of the sides}) \). Substituting the areas gives: \[ C = 4 \times 2\pi x^2 + 1 \times \frac{80}{x} = 8\pi x^2 + \frac{80}{x}. \]

Step 4: Find Derivative of Cost Function

To find the radius that minimizes cost, take the derivative of the cost function \( C \) with respect to \( x \): \[ C'(x) = 16\pi x - \frac{80}{x^2}. \]

Step 5: Set Derivative to Zero

Set the derivative equal to zero to find critical points: \[ 16\pi x - \frac{80}{x^2} = 0. \] Solving for \( x \): \[ 16\pi x^3 = 80 \Rightarrow x^3 = \frac{80}{16\pi} \Rightarrow x = \left(\frac{5}{\pi}\right)^{1/3}. \]

Step 6: Use a Calculator for Exact Radius

Compute \( x = \left(\frac{5}{\pi}\right)^{1/3} \) using a calculator to find the exact value. This provides the radius for minimum cost.

Key Concepts

Rational Functions

Rational functions are a type of function represented as the ratio of two polynomials. They are central in modeling situations where quantities depend on each other in multiplicative or divisive ways. When dealing with practical problems, such as optimizing costs or designing structures, rational functions can describe relationships between dimensions and constraints. In our problem, a rational function helps us express both the cylinder’s surface area and cost in terms of the radius.

• The numerator of a rational function describes one quantity (e.g., area or cost).
• The denominator represents another aspect influencing the first (e.g., radius).

Calculating with rational functions usually involves finding critical points, as these indicate where the function has maximum or minimum values. In this exercise, the critical point found by setting the derivative equal to zero provides the radius that minimizes the construction cost.

Cylinder Volume

The volume of a right circular cylinder is a core geometric formula, often expressed as \( V = \pi x^2 h \). Here, \( x \) is the radius of the base circle, and \( h \) is the height. This formula arises from the fact that the volume is essentially the area of the circle base times the cylinder’s height. Knowing how to rearrange this formula allows for solving one variable in terms of others.

• Volume is given (e.g., 40 cubic inches in this exercise).
• You must often solve for one variable, like height, after assigning a value for another, like radius.

In our exercise, setting the volume equal to 40 and solving for \( h \) in terms of \( x \) ensures that any chosen radius fits this volume constraint. This interdependence is crucial when minimizing cost while meeting structural specifications.

Surface Area Calculations

Understanding and calculating the surface area of a cylinder involves dealing with both flat and curved surfaces. An essential formula for total surface area, \( A \), considers the top and bottom circles and the curved side: \( A = 2\pi x^2 + 2\pi x h \).

• The top and bottom areas together are \( 2\pi x^2 \).
• The lateral or side area is a rectangle that, when unrolled, has area \( 2\pi x h \).

By substituting \( h \) from the volume equation, you can express the surface area entirely in terms of \( x \), allowing for direct analysis concerning radius. The calculation of surface area is a necessary step before converting these physical properties into a monetary cost function for optimization.

Cost Optimization

Cost optimization involves determining how to achieve the best financial outcome under specific constraints. In the cylinder problem, we minimize the total construction cost by developing a cost function that includes materials priced differently for different surfaces: top and bottom versus the side of the cylinder.

• The cost function, \( C = 8\pi x^2 + \frac{80}{x} \), depends on surface areas affected by the given costs per square inch.
• By differentiating this cost function with respect to \( x \), we identify points at which cost changes stop, helping define critical values.

Setting the derivative to zero reveals these critical points, crucial for finding the optimal radius to minimize costs. Exact numerical solutions using calculators provide practical implementations, ensuring precise construction specifications.