Question

Of all right circular cylinders with a given surface area, find the one with the maximum volume. Note: The ends of the cylinders are closed.

Short answer

The cylinder with maximum volume has equal height and radius, \(r = h = \sqrt{\frac{S}{6\pi}}\).

Step-by-step solution

Understand the Problem

We want to find the right circular cylinder with a given surface area that maximizes volume. The surface area of a right cylinder with radius \(r\) and height \(h\) is given by \(2\pi r^2 + 2\pi rh = S\), and the volume is \(V = \pi r^2 h\). Our task is to express the volume in terms of surface area, and maximize it.

Express Height in Terms of Radius

From the surface area formula, \(2\pi r^2 + 2\pi rh = S\), solve for \(h\): \(h = \frac{S - 2\pi r^2}{2\pi r}\). This expresses the height \(h\) as a function of radius \(r\).

Express Volume in Terms of Radius

Substitute the expression for \(h\) from Step 2 into the volume formula \(V = \pi r^2 h\): \(V = \pi r^2 \left(\frac{S - 2\pi r^2}{2\pi r}\right)\). Simplifying gives \(V = \frac{r}{2}(S - 2\pi r^2)\).

Differentiate Volume with respect to Radius

To find the maximum volume, differentiate \(V\) with respect to \(r\): \(\frac{dV}{dr} = \frac{S}{2} - 3\pi r^2\).

Set the Derivative to Zero and Solve for Radius

Set the derivative equal to zero to find critical points: \(\frac{S}{2} - 3\pi r^2 = 0\). Solving gives \(r^2 = \frac{S}{6\pi}\), and therefore \(r = \sqrt{\frac{S}{6\pi}}\).

Find Corresponding Height for Maximum Volume

Substitute \(r = \sqrt{\frac{S}{6\pi}}\) back into the formula for \(h\): \(h = \frac{S - 2\pi r^2}{2\pi r}\). Simplifying gives \(h = \sqrt{\frac{S}{6\pi}}\) as well.

Analyze Results

The maximum volume occurs when the cylinder's height equals its radius: \(h = r\). Thus, the right circular cylinder with the maximum volume given a fixed surface area is a cylinder where the height equals the radius.

Key concepts

Right Circular Cylinder

A right circular cylinder is a three-dimensional geometric shape characterized by a base that is a circle with two parallel circular ends and a curved surface connecting them. This shape is commonly used in many real-world applications such as in cans and storage tanks.
The cylinder's geometry is quite straightforward: it has a radius, denoted by \(r\), and a height, \(h\). Its volume \(V\) is given by the formula \(V = \pi r^2 h\) which represents the space inside the cylinder. The surface area \(S\) includes both the curved surface and the circles at the top and bottom, calculated as \(S = 2\pi r^2 + 2\pi rh\).
The study of such a cylinder, particularly with closed ends, is fundamental in optimization problems often seen in calculus and is essential for understanding how these shapes can be created and manipulated in various fields.

Volume Maximization

Maximizing the volume of a right circular cylinder with a given surface area is a common optimization problem in calculus. The goal is to find dimensions that yield the greatest volume possible while keeping the total surface area constant. This is particularly important in many engineering and design applications where the efficiency of space use is crucial.
To find the maximum volume, we must express the volume \(V\) in terms of one variable, as given in the formula \(V = \frac{r}{2}(S - 2\pi r^2)\). This transformation allows us to explore how changes in radius \(r\) might affect the volume while keeping the surface area \(S\) constant.
Finding the optimal radius and height to achieve maximum volume is key here since engineers and designers often face constraints in the materials they use, dictating specific maximum surface areas for construction.

Surface Area Constraints

Surface area constraints play a significant role in optimizing the volume of a cylinder. They represent a real-life limitation where there is a fixed amount of material available to create the cylinder. The constraint is expressed mathematically as \(S = 2\pi r^2 + 2\pi rh\), the total surface area of the cylinder.
By working within this constraint, we determine how to allocate the surface area effectively between the base and the height to maximize the internal volume. This involves solving equations that relate to these different components of the surface area to the desired result of maximizing the cylinder's volume.
Solving optimization problems under such constraints helps improve efficiency and resource utilization, whether in manufacturing, packaging, or storage facilities.

Derivative Applications

Derivatives are powerful tools in calculus used to find maximum or minimum values of a function, which in this case, is used to maximize the volume of a cylinder. When we take the derivative of the volume equation \(V = \frac{r}{2}(S - 2\pi r^2)\) with respect to the radius \(r\), we obtain \(\frac{dV}{dr} = \frac{S}{2} - 3\pi r^2\).
Setting the derivative to zero helps find the critical points where the volume could be maximized. Solving \(\frac{S}{2} - 3\pi r^2 = 0\), we find the optimal radius \(r = \sqrt{\frac{S}{6\pi}}\).
This use of derivatives is a practical application that allows us to not only find these points quickly but also to confirm whether these critical points achieve the desired maximum through further analysis, such as second derivative tests.