Problem 14 - Calculus: Early Transcendentals

Question

You want to make cylindrical containers of a given volume \(V\) 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 optimal ratio of height to radius.

Step-by-Step Solution

Answer

The optimal height to radius ratio is \(\frac{4}{\pi^{1/3}}\).

Step 1: Understand the problem with constraints

You are asked to find the ratio of height to radius for a cylindrical container that minimizes the amount of material needed. The prescribed volume of the cylinder is given as \( V \), and the material needed for the top and bottom is calculated differently than standard formulas.

Step 2: Define the volume constraint equation

The volume \( V \) of a cylinder with radius \( r \) and height \( h \) is given by the formula \( V = \pi r^2 h \). This equation will relate \( h \) and \( r \) under the given volume constraint.

Step 3: Set up the surface area function to minimize

The amount of material used is the sum of the lateral area (\( 2\pi rh \)) and the special top and bottom area (given as \( 8r^2 \)). Therefore, the surface area function \( S \) to minimize is \( S = 2\pi rh + 8r^2 \).

Step 4: Solve for height in terms of radius

From the volume constraint \( V = \pi r^2 h \), solve for \( h \) to get \( h = \frac{V}{\pi r^2} \).

Step 5: Substitute height into the surface area function

Substitute \( h = \frac{V}{\pi r^2} \) into \( S = 2\pi rh + 8r^2 \) to express \( S \) only in terms of \( r \). This results in \( S = 2\pi r\left(\frac{V}{\pi r^2}\right) + 8r^2 = \frac{2V}{r} + 8r^2 \).

Step 6: Find the critical points

To minimize \( S \), take its derivative with respect to \( r \) and set it to zero: \( \frac{dS}{dr} = -\frac{2V}{r^2} + 16r = 0 \). Solve this equation for \( r \).

Step 7: Solve for the optimal radius

Rearrange \( -\frac{2V}{r^2} + 16r = 0 \) to \( 16r = \frac{2V}{r^2} \). Solving gives \( r^3 = \frac{V}{8} \), so \( r = \left(\frac{V}{8}\right)^{1/3} \).

Step 8: Calculate the height-to-radius ratio

With \( r = \left(\frac{V}{8}\right)^{1/3} \), substitute back into \( h = \frac{V}{\pi r^2} \) to get \( h = \frac{V}{\pi \left(\frac{V}{8}\right)^{2/3}} = \frac{8^{2/3} V^{1/3}}{\pi} \). The ratio of height to radius is then \( \frac{h}{r} = \frac{8^{2/3}}{\pi} \).

Step 9: Simplify the ratio

This simplifies to \( \frac{h}{r} = \frac{4}{\pi^{1/3}} \). Therefore, the optimal height to radius ratio is \( \frac{4}{\pi^{1/3}} \).

Key Concepts

Cylindrical Containers

Cylindrical containers are an interesting problem in optimization because of their practical applications in manufacturing and packaging. When constructing these containers, the goal is often to use the least amount of material possible. Here, we focus on a cylinder which is a 3-dimensional shape with a circular base and a specific height. In this exercise, we're tasked with optimizing the surface area of these cylinders.

The exercise specifically highlights a scenario where material use is minimized not only for the lateral surface but also for the unique way the top and bottom are created. Instead of the typical circular top and bottom, material efficiency is achieved by using squares from which circles are intended, leading to a greater material usage than a simple circle. Understanding this distinction is essential as it sets different conditions for finding the optimal design.

Volume Constraint

A key part of understanding optimization in cylindrical containers is the concept of a volume constraint. The volume of a cylinder is calculated using the formula \( V = \pi r^2 h \), where \( r \) is the radius of the base, and \( h \) is the height.

This relationship implies that if we know the volume, which is often a fixed requirement, we can derive various relationships between \( r \) and \( h \). In our problem, the volume is predetermined, serving as a constant. This constraint is critical because it guides how changes in radius and height will maintain the required volume, allowing us to explore how to adjust the dimensions while minimizing the surface area.

Surface Area Minimization

Minimizing the surface area of a cylinder involves creating an equation for the total material used. In this problem, the lateral surface is combined with the surface area of the top and bottom. The surface area function \( S \) is therefore a sum of the lateral area \( 2\pi rh \) and the area due to the special top and bottom treatment, \( 8r^2 \).

To minimize \( S \), it needs to be expressed in terms of one variable. By substituting the expression of \( h \) derived from the volume constraint, the surface area function can be simplified to \( S = \frac{2V}{r} + 8r^2 \). This formulation is critical because it allows the use of calculus to determine the optimal dimensions.

Derivative Application

Calculus provides powerful tools for solving optimization problems, particularly through the use of derivatives. By taking the derivative of the surface area equation \( S = \frac{2V}{r} + 8r^2 \) with respect to \( r \), you can find the critical points which indicate potential minima or maxima of the function.

Setting the derivative \( \frac{dS}{dr} = -\frac{2V}{r^2} + 16r \) to zero isolates the values of \( r \) where the surface area could be minimized. Solving the resulting equation leads to the expression \( r^3 = \frac{V}{8} \). Taking the cube root gives the optimal radius, \( r = \left(\frac{V}{8}\right)^{1/3} \). With this, substituting back into derived expressions yields the optimal height-to-radius ratio, which is the ultimate goal of the problem.