Problem 56 - Calculus Single Variable

Question

Exercises \(56-58\) concern the most economical dimensions of a cylindrical tin can that holds a predetermined volume \(V\). What are the dimensions if the amount of metal used in the can is to be minimized? Express your answer as a height \(h\) to radius \(r\) ratio.

Step-by-Step Solution

Answer

The height to radius ratio is 2:1.

Step 1: Understand the Problem

We need to minimize the surface area of a cylindrical can with a fixed volume while determining the ratio of the height to the radius.

Step 2: Define Variables and Formulas

The volume of the cylinder is given by \( V = \pi r^2 h \) and the surface area \( A \) is given by \( A = 2\pi rh + 2\pi r^2 \). Our goal is to minimize \( A \) while keeping \( V \) constant.

Step 3: Express Height in Terms of Volume and Radius

Since \( V = \pi r^2 h \), we can express \( h \) as \( h = \frac{V}{\pi r^2} \).

Step 4: Substitute Height in Surface Area Formula

Replace \( h \) in the surface area formula: \( A = 2\pi r \left(\frac{V}{\pi r^2}\right) + 2\pi r^2 \) which simplifies to \( A = \frac{2V}{r} + 2\pi r^2 \).

Step 5: Find the Derivative of the Surface Area Function

Differentiate \( A = \frac{2V}{r} + 2\pi r^2 \) with respect to \( r \): \( \frac{dA}{dr} = -\frac{2V}{r^2} + 4\pi r \).

Step 6: Set the Derivative to Zero and Solve for r

Find \( r \) by setting the derivative to zero: \( -\frac{2V}{r^2} + 4\pi r = 0 \). Solving this gives \( 4\pi r^3 = 2V \), so \( r^3 = \frac{V}{2\pi} \) and \( r = \left(\frac{V}{2\pi}\right)^{1/3} \).

Step 7: Calculate Height Using Radius

Substitute \( r = \left(\frac{V}{2\pi}\right)^{1/3} \) into the equation for \( h = \frac{V}{\pi r^2} \), resulting in \( h = \frac{V}{\pi \left(\frac{V}{2\pi}\right)^{2/3}} \). Simplifying gives \( h = 2 \left(\frac{V}{2\pi}\right)^{1/3} \).

Step 8: Determine Height to Radius Ratio

The height-to-radius ratio \( \frac{h}{r} \) is \( \frac{2 \left(\frac{V}{2\pi}\right)^{1/3}}{\left(\frac{V}{2\pi}\right)^{1/3}} = 2 \).

Key Concepts

Cylindrical Can Dimensions

When it comes to designing a cylindrical can, the dimensions we're primarily concerned with are the height () and the radius () of the can's base. The volume of a cylindrical can is calculated using the formula:

• \( V = \pi r^2 h \)

Here, \( V \) is the volume, \( r \) is the radius, and \( h \) is the height. This formula helps ensure that the can will hold a specific amount of material, such as a beverage or food item, without spilling over.
To move forth with any optimization, understanding the relationship between these variables is essential. As the radius and height change, they affect both the volume and the surface area of the can. The goal is to find the most economical configuration where the can still holds the desired volume while using the least material to make it.

Surface Area Minimization

Minimizing the surface area of the cylindrical can implies using the least amount of material while maintaining the necessary volume. The surface area () of a cylindrical can is given by the formula:

• \( A = 2\pi rh + 2\pi r^2 \)

This formula accounts for the lateral area of the sides (\( 2\pi rh \)) and the area of the two circular ends (\( 2\pi r^2 \)).
To achieve minimization, the idea is to solve for \( h \) in terms of \( V \) and \( r \), and then substitute it back into the surface area formula. This substitution ultimately gives rise to an equation in terms of \( r \) alone.
During this process, we compute the derivative of the surface area function with respect to \( r \) and set it to zero. This step is crucial as it helps us determine the optimal radius value that minimizes the surface area for the given volume. Solving this derivative equation offers insights into the balance between the height and the radius that results in the least material usage for the can.

Volume Constraints

Constraints on volume come into play when the volume of the cylindrical can is predetermined or fixed. This means that the cylinder must be designed to hold an exact amount of material, which directly informs how we choose our dimensions.

• For instance, if we hold the volume constant as \( V \), it implies we need to ensure this constraint is met throughout all our calculations.

We achieve this by expressing the height in terms of the volume and radius:

• \( h = \frac{V}{\pi r^2} \)

This relationship helps lock the volume constraint into our equations for the surface area minimization.
Carefully adhering to volume constraints is vital, not only to ensure functionality of the can but also to make sure that material and cost efficiency is optimized without compromising on the intended capacity. Whenever changes to the dimensions occur for optimization, the volume constraint serves as a check to ensure the result still satisfies the intended purpose of holding a specific volume.