Problem 14 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

A can for mandarin oranges is a cylinder with volume of 250 cubic centimeters. Denote the radius by \(r\) and the height by \(h .\) The material used for the top and bottom is stronger than that used for the sides. There is wasted material in constructing the top and bottom because they need to be cut from squares of metal and the scrap metal is not used. The manufacturers must pay for the material for the whole square from which the circle is cut. Suppose that the material for the top and bottom is three times as expensive as the material for the sides. What are the dimensions of the can that minimize the cost of the materials?

Step-by-Step Solution

Answer

The dimensions that minimize the cost are a radius of \(\sqrt[3]{\frac{125}{6\pi}}\) cm and a height of \(2\sqrt[3]{\frac{125}{6\pi}}\) cm.

Step 1: Determine the Qualitative Relationship Between Variables

The volume of the cylinder is given by \(V = \pi r^2 h\). We are told this equals 250 cm³. Rearranging for \(h\) gives \(h = \frac{250}{\pi r^2}\).

Step 2: Define the Cost Function

We are told that the material in the top and bottom (two circles) is three times as expensive than the side (a rectangle unwrapped). The cost function \C(r)\ is thus \C(r) = 3 * 2 * \(\pi r^2\) + \(\pi 2*r*h\). Using \(h = \frac{250}{\pi r^2}\) from Step 1, this simplifies to \C(r) = 6\(\pi r^2\) + \(\frac{500}{r}\).

Step 3: Optimize the Cost Function

To find the minimum cost, we should find the derivative of the cost function and set it equal to zero, then solve for the radius \(r\). So \C'(r) = 12\(\pi r\) - \(\frac{500}{r^2}\) = 0. Solving this equation for \(r\) we get \(r = \(\sqrt[3]{\frac{125}{6\pi}}.\)

Step 4: Find the value of h

Once we have the value of \(r\), we can find \(h\) using the relation from Step 1. Substituting the value of \(r\) into \(h = \frac{250}{\pi r^2}\) gives \(h = 2\sqrt[3]{\frac{125}{6\pi}}.\)

Key Concepts

Volume of a Cylinder

The volume of a cylinder is an essential concept in geometry, particularly useful in various applications, including this real-life problem involving the optimization of material costs for manufacturing a can. The volume, represented by the formula \(V = \text{π}r^2h\), denotes the amount of space inside the cylinder. Here, \(r\) stands for the radius of the cylinder's base, and \(h\) represents its height. For a can designed to hold 250 cubic centimeters, we can manipulate this formula to express either radius or height in terms of the other, enabling us to analyze how changing one dimension affects the volume and the cost.

Understanding this relationship is vital. By adjusting the radius and height while keeping the volume constant, we can explore different can shapes that could reduce the cost of materials needed for production, while still meeting the volume requirement. Calculating the volume is the first step in our journey to cost optimization.

Cost Function Optimization

Cost function optimization is a cornerstone of business operations and economics. In our context, it involves finding the dimensions of a can that will minimize the material costs. The cost function \(C(r)\) takes into account the different expenses associated with the materials for the can's top, bottom, and sides, recognizing that the top and bottom are subject to additional waste due to the manufacturing process.

Setting up the cost function correctly is crucial, as it reflects the reality of production costs and also provides a mathematical expression that can be manipulated using calculus to find the optimal solution. The cost function here combines the cost of the circular tops and bottoms with that of the rectangular side, and by incorporating the volume constraint, we create a single-variable function that can be optimized using derivatives. This blending of geometry, economics, and calculus provides a powerful tool for businesses to minimize costs and maximize efficiency.

Derivatives in Calculus

Derivatives are at the heart of calculus, serving as a fundamental tool for understanding how quantities change. In the quest to minimize the can's cost, we employ derivatives to determine how the cost function behaves when we vary the radius of the cylinder. Taking the derivative of the cost function with respect to the radius, \(C'(r)\), provides us with the rate at which the cost changes as the radius changes.

When we set this derivative equal to zero, we're looking for where the cost function has a critical point, which could indicate a minimum cost in this context. By finding the roots of the derivative, we identify the radius that yields the least costly can design. This application of derivatives is a prime example of how calculus enables us to solve practical optimization problems by providing a systematic method to analyze and identify optimal solutions.

Application of Calculus in Economics

Calculus is an indispensable tool in economics for solving optimization problems like the one we're examining. Its principles allow economists and business leaders to make informed decisions about cost efficiency. By applying the techniques of calculus to economic functions – such as the cost function in our exercise – one can determine the most economically viable solutions, like the proportions of a can that minimize material costs.

As we've seen, by translating an economic situation into a mathematical model, evaluating the model using calculus, and interpreting the results, we unlock powerful insights. These insights can lead to significant cost savings and improved resource utilization, demonstrating the profound impact of calculus on decision-making in economics. The union of calculus and economics thus paves the way for innovations in cost management and other economic phenomena. In education, these real-world applications help bridge the gap between abstract mathematical concepts and practical, tangible outcomes.