Problem 88 - College Algebra with Corequisite Support

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 to yield minimum cost is approximately 1.84 inches.

Step 1: Identify Given Values

First, recognize the given values and the information we need to work with. We know the volume of the cylinder is 40 cubic inches, and that the cost for materials depends on the surface area of the cylinder. The cost is 4 cents per square inch for the top and bottom, and 1 cent per square inch for the sides.

Step 2: Write Formula for Volume

The volume, V, of a cylinder is given by the formula: \[ V = \pi x^2 h \]where \(x\) is the radius and \(h\) is the height of the cylinder. Given the volume is 40 cubic inches, we have:\[ \pi x^2 h = 40 \]

Step 3: Solve for Height

Rearrange the volume formula to solve for the height \(h\):\[ h = \frac{40}{\pi x^2} \]

Step 4: Calculate Surface Areas

The surface area of the top and bottom (two circles) is: \[ A_{top\&bottom} = 2\pi x^2 \]The surface area of the side (a rectangle) is:\[ A_{side} = 2\pi x h \]Substitute \(h = \frac{40}{\pi x^2} \) into \(A_{side}\):\[ A_{side} = 2\pi x \left(\frac{40}{\pi x^2}\right) = \frac{80}{x} \]

Step 5: Express Cost Function

The cost, \(C\), of constructing the cylinder is determined by multiplying each surface area with their respective construction costs:\[ C = 4(2\pi x^2) + 1\left(\frac{80}{x}\right) \]Simplify to:\[ C = 8\pi x^2 + \frac{80}{x} \]

Step 6: Find the Derivative and Critical Points

To find the radius that minimizes cost, we need to find the critical points of \(C\):Find the derivative of \(C\) with respect to \(x\):\[ C'(x) = 16\pi x - \frac{80}{x^2} \]Set \( C'(x) = 0\) to find critical points:\[ 16\pi x = \frac{80}{x^2} \]Multiply through by \(x^2\) and simplify:\[ 16\pi x^3 = 80 \Rightarrow x^3 = \frac{80}{16\pi} \]

Step 7: Calculate Optimal Radius

Solve for \(x\):\[ x = \left(\frac{80}{16\pi}\right)^{1/3} \]Use a calculator to approximate the value of \(x\).

Step 8: Verify and Conclude

Check the second derivative or use another method to confirm the critical point is a minimum. Given the structure of the problem, evaluating beyond the first derivative may not be explicitly necessary once finding\(x\). The optimal radius minimizes the cost of constructing the cylinder.

Key Concepts

Cylinder Surface Area

Understanding the surface area of a cylinder is crucial when dealing with material costs, especially in practical problems like constructing a cylinder. A cylinder's surface area consists of two parts: the top and bottom circles, and the side, which unfolds into a rectangle.

• Top and Bottom Areas: Each circular area is given by the formula \ A = \pi x^2 \, where \( x \) is the radius. Since there are two such circles in a cylinder, the total area for the top and bottom is \ 2\pi x^2 \.
• Side Area: This is calculated by visualizing the side as a rectangle with height \( h \) and a width equal to the cylinder's circumference, \ 2\pi x \. Thus, the side area is given by \ A_{side} = 2\pi x h \.

These areas are used to determine the material needed, impacting the cost when constructing the cylinder.

Cost Optimization

In problems similar to this one, cost optimization can mean significant savings. Here, the main goal is to find the radius that minimizes the total cost of constructing the cylinder.

• The cost associated with the top and bottom areas is 4 cents per square inch.
• The cost for the side is 1 cent per square inch.
• The total cost \( C \) can be expressed as a function of the radius \( x \).
• This function, based on the calculated surface areas, is \ C = 8\pi x^2 + \frac{80}{x} \.

Using this cost function, we aim to find a particular radius that results in the lowest possible cost. This is where calculus tools, like finding derivatives, become essential.

Derivatives and Critical Points

Derivatives are instrumental in finding maxima and minima for optimization problems. Here’s how they’re applied in finding the minimum cost for constructing the cylinder:

• First, find the derivative \( C'(x) \) of the cost function \( C(x) = 8\pi x^2 + \frac{80}{x} \).
• The derivative is calculated as \( C'(x) = 16\pi x - \frac{80}{x^2} \).
• Critical points, where the function achieves its min or max values, occur when the derivative equals zero: \( 16\pi x = \frac{80}{x^2} \).
• Set this equation and solve for \( x \), which provides the radius giving the minimum cost.

Critical points are fundamental in optimization because they help reveal where potential minimum or maximum values of the function occur. Solving them requires rearranging and simplifying equations.

Volume of a Cylinder

The volume of a cylinder is defined as the amount of space inside the cylinder. The formula for the volume \( V \) of a cylinder is \ V = \pi x^2 h \.

• Here, \( x \) is the radius and \( h \) is the height of the cylinder.
• For this specific problem, we know the volume is constrained to 40 cubic inches, giving us an equation. \ \pi x^2 h = 40 \.
• This relation helps us find expressions for the height \( h \) in terms of \( x \), which is necessary for further calculations.

Understanding the volume formula helps interpret parameters defining a cylinder's dimensions and aids in deriving other essential properties, like its surface area for cost calculations.