Problem 29 - Calculus Early Transcendentals

Question

A closed, cylindrical can is to have a volume of \(V \mathrm{cubic }\)units. Show that the can of minimum surface area is achieved when the height is cqual to the diameter of the base.

Step-by-Step Solution

Answer

The can achieves a minimum surface area when the height equals the diameter.

Step 1: Understand the Problem and the Formula

We need to find the dimensions of a cylindrical can with a minimum surface area for a given volume \( V \). The surface area of a cylinder consists of the side (which is a rectangle when unrolled) and the two circular bases.

Step 2: Define the Known Parameters

Let the radius of the base be \( r \) and the height be \( h \). The volume \( V \) is given by \( \pi r^2 h = V \). We also need the surface area \( A \) which is \( 2\pi r^2 + 2\pi rh \).

Step 3: Express Height in Terms of Radius and Volume

Using the volume formula, express height \( h \) in terms of radius \( r \) and volume \( V \): \( h = \frac{V}{\pi r^2} \).

Step 4: Formulate the Surface Area in Terms of Radius

Substitute \( h \) from Step 3 in the surface area formula: \( A = 2\pi r^2 + 2\pi r \left(\frac{V}{\pi r^2}\right) = 2\pi r^2 + \frac{2V}{r} \).

Step 5: Differentiate the Surface Area with Respect to Radius

Find the derivative of \( A \) with respect to \( r \): \( \frac{dA}{dr} = 4\pi r - \frac{2V}{r^2} \).

Step 6: Identify Critical Points for Minimum Surface Area

Set \( \frac{dA}{dr} = 0 \) to find critical points: \( 4\pi r - \frac{2V}{r^2} = 0 \). Rearrange this to find \( r \): \( 4\pi r^3 = 2V \), and hence \( r^3 = \frac{V}{2\pi} \).

Step 7: Calculate Maximum or Minimum Volume Condition

Knowing \( r^3 = \frac{V}{2\pi} \), solve for \( r \): \( r = \left(\frac{V}{2\pi}\right)^{1/3} \). Substitute back into the expression for \( h \) to find \( h = \frac{V}{\pi r^2} = 2 \left(\frac{V}{2\pi}\right)^{1/3} = 2r \).

Step 8: Conclusion Based on Calculations

From the calculations, when \( h = 2r \), the height is equal to the diameter. Thus, the can achieves the minimum surface area when the height is equal to the diameter.

Key Concepts

Cylindrical Can Surface Area

The problem involves minimizing the surface area of a cylindrical can with a fixed volume. To understand this in simpler terms, think of the can as having three parts: a side (like a label around a can) and two circular ends (the top and bottom). The surface area is influenced by both the radius of the base and the height of the can.
This means our task is to find a balance between these two measurements so that the can uses the smallest amount of material possible. The surface area formula captures this balance and is expressed as \(2\pi r^2 + 2\pi r h\), where \(r\) is the radius and \(h\) is the height. The challenge lies in adjusting \(r\) and \(h\) while keeping the volume constant to minimize the overall surface area.

Derivative and Critical Points

To find out how to adjust the can’s dimensions for a minimum surface area, we use calculus, specifically derivatives. A derivative helps us understand how a function changes as its input changes. Here, our function is the surface area formula rewritten in terms of radius. We differentiate this formula to find the rate of change of surface area concerning radius.
By setting the derivative equal to zero \(\frac{dA}{dr} = 0\), we identify the critical points. These points are potential candidates for maxima or minima, where the surface changes its increasing or decreasing behavior. Solving \(4\pi r - \frac{2V}{r^2} = 0\) gives us a critical point which provides a specific radius that could result in the minimum surface area.

Volume and Surface Area Formulas

Understanding the key formulas is crucial in solving optimization problems. The volume of the cylindrical can is given by the formula \(V = \pi r^2 h\), representing the space enclosed within the can. The surface area formula, \(A = 2\pi r^2 + 2\pi r h\), calculates the total area covering the can.
When performing optimization, we aim to rewrite such formulas so that we only need to deal with one variable, in this case, radius \(r\). By expressing height \(h\) in terms of \(r\), we simplify the problem, reducing the complexity and making it easier to apply calculus techniques.

Mathematical Proofs

Mathematical proofs are logical arguments demonstrating why certain mathematical statements are true. In this problem, after calculating that \(h = 2r\), we prove the statement that the can achieves minimum surface area when its height equals the diameter of its base.
We confirm this through solving the critical point equations and ensuring that these conditions align back with our formulas. Verifying the relationship \(h = 2r\) completes the proof, showcasing how mathematical reasoning and calculations collaborate to prove a theorem or stave a conjecture.