Problem 13 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

A tin can for garbanza beans is designed to be a cylinder with volume of 300 cubic centimeters. Denote the radius by \(r\) and the height by \(h\). The top and bottom are thicker than the sides; for the purposes of our model, we'll assume that they are made with a double thickness of aluminum. (a) Give an expression for the volume of the can. (b) Give an expression for the amount of material used. (Remember that the top and bottom of the can are two layers thick.) (c) Make the expression from part (b) into a function of \(r\) alone. (d) What radius minimizes the material used? (e) What are the dimensions of the 300 cubic-centimeter can that require the least amount of material?

Step-by-Step Solution

Answer

The 300 cubic-centimeter can that requires the least amount of material has a radius of \(r = \sqrt[3]{\frac{75}{\pi}}\) cm and a height of \(h = 2 \sqrt[3]{\frac{75}{\pi}}\) cm.

Step 1: Expression for Volume

The volume \(V\) of the cylinder can be given by \(V = \pi r^2 h\). According to the problem we know that \(V=300\), so expressing \(h\) we get \(h = \frac{300}{\pi r^2}\)

Step 2: Expression for Material Used

The surface area of a cylinder is given by \(2\pi r (r + h)\). However, the problem states that the top and bottom of the can are two layers thick, so we need to account for this. The expression for the material used \(A\) therefore becomes \(A = 2(2\pi r^2) + 2\pi r h = 4\pi r^2 + 2\pi r h\)

Step 3: Function of r Alone

Substitute \(h\) from Step 1 into the expression for \(A\) from Step 2. So, the function \(A(r)\) becomes \(A(r) = 4\pi r^2 + \frac{600}{r}\)

Step 4: Calculate Derivative and Find Minimum

To find the radius that minimizes the amount of material used, take the derivative of \(A(r)\) and set it to zero. So, \(A'(r) = 8\pi r - \frac{600}{r^2} = 0\), simplifying this we get \(r= \sqrt[3]{\frac{75}{\pi}}\)

Step 5: Find h and Verify Solution

Substitute the value of r into the equation for h in Step 1, then check the second derivative of \(A(r)\) to make sure it's positive to demonstrate that it is indeed a minimum. Hence on substituting we get \(h = 2 \sqrt[3]{\frac{75}{\pi}}\). The second derivative of \(A(r)\) is \(A''(r) = 8\pi + \frac{1200}{r^3}\), substituting r value we find \(A''(r)\) is positive

Step 6: Finalize Dimensions

The dimensions of the can that minimize the amount of material used, with a volume of 300 cubic centimeters, are a radius of \(r = \sqrt[3]{\frac{75}{\pi}}\) cm and height \(h = 2 \sqrt[3]{\frac{75}{\pi}}\) cm.

Key Concepts

Cylindrical Shapes

When considering optimization problems involving volumes and surface areas, it's often helpful to start by understanding the shapes involved. A cylindrical shape, like the tin can in our exercise, is formed by two parallel circular bases connected by a curved surface. The properties of cylinders include parameters such as:

• Radius ( \( r \) ): the distance from the center of the base to its edge.
• Height ( \( h \) ): the distance between the two bases.

Understanding these components is key, as they play a crucial role in calculating both the volume and the material needed to construct such shapes. Cylinders are ubiquitous in many real-world applications, from containers to machinery parts, making them essential in both geometry and practical problem-solving.

Volume and Surface Area Formulas

To solve problems involving cylindrical shapes, you need to know some basic formulas.The volume (\( V \) ) of a cylinder is given by:\[V = \pi r^2 h\]where \( \pi \)is a mathematical constant approximately equal to 3.14159. This formula expresses the space inside the cylinder.The surface area (\( A \) ) includes both the sides and the top and bottom surfaces. The general surface area for a basic cylinder is:\[A = 2times \pi r^2 + 2 times \pi r h\]In cases like this problem, where the top and bottom are two layers thick, adjustments are necessary.Here, the correct expression for material used is:\[A = 4times \pi r^2 + 2 imes \pi r h\]Applying these formulas effectively is fundamental in calculating how much material a project requires.

Derivative Test for Minimum

To find optimal solutions, such as minimizing the material used for the can, calculus offers powerful tools. One of these is the derivative test, which helps identify minimum (or maximum) values of a function.In this exercise, once we transform the surface area function into a function of \( r \) alone, we use the derivative:\[A'(r) = 8times \pi r - \frac{600}{r^2}\]Setting \( A'(r) = 0 \) allows us to find the critical points that potentially minimize material usage.After finding the critical radius, using the second derivative, \( A''(r) \), confirms whether this point is a minimum. If \( A''(r) > 0 \), it indicates a local minimum, meaning the material used is indeed minimized at the computed radius. Calculating accurately and interpreting these results are key steps in solving optimization challenges.

Material Minimization

Material minimization is a common goal in design and manufacturing. It involves creating products that use the least amount of materials while fulfilling their intended purpose.For the cylindrical can problem, minimizing material affects both economic and environmental concerns. The derived optimal dimensions, \( r \) and \( h \), ensure that the least amount of aluminum is used. Utilizing strategies like the one in this problem, where both volume constraints and material constraints are considered, leads to efficient and sustainable design practices. Being proficient in applying these concepts allows for the efficient design of items beyond just cans, helping minimize costs and material waste in industries ranging from packaging to engineering.