Problem 19 - Calculus for Biology and Medicine

Question

Optimal Soda Can A soda can manufacturer wants to minimize the cost of the aluminum used to make their can. The can has to hold a volume \(V\) of soda. Assuming that the thickness of the can is the same everywhere, the amount of aluminum used to make the can will be proportional to its surface area. That is, suppose the height of the can is \(h\) and the radius of the can is \(r\), as in Figure \(5.56 .\) Then the manufacturer wants to minimize: $$ S=2 \pi r h+2 \pi r^{2} $$ subject to the constraint that \(\pi r^{2} h=V .\) Here we have used the formulas for the total surface area and volume of a cylinder. (a) A real soda can has volume \(V=355 \mathrm{~cm}^{3}\) (or \(12 \mathrm{fl}\). oz.). By substituting for \(h\) in Equation \((5.17)\), write \(S\) as a function of \(r\) only. (b) Describe the behavior of \(S(r)\) as \(r \rightarrow \infty\) (c) Describe the behavior of \(S(r)\) as \(r \rightarrow 0\). (d) Based on your answers to (b) and (c), explain why you expect there to be a value of \(r\) that minimizes \(S(r) .\) Calculate this optimum radius \(r\).

Step-by-Step Solution

Answer

The optimal radius to minimize the aluminum used is approximately 3.84 cm.

Step 1: Express Height in Terms of Radius

We're given the constraint for volume: \[ \pi r^2 h = V \]Solving for the height \( h \), we get: \[ h = \frac{V}{\pi r^2} \] .

Step 2: Substitute for Height in Surface Area Formula

Now substitute \( h = \frac{V}{\pi r^2} \) into the surface area formula:\[ S = 2 \pi r \left( \frac{V}{\pi r^2} \right) + 2 \pi r^2 \]Simplifying, we get:\[ S(r) = \frac{2V}{r} + 2\pi r^2 \] .

Step 3: Analyze Behavior of Surface Area as \( r \rightarrow \infty \)

As \( r \) approaches infinity, the term \( \frac{2V}{r} \) approaches zero and \( 2\pi r^2 \) grows without bound. Therefore, \( S(r) \rightarrow \infty \).

Step 4: Analyze Behavior of Surface Area as \( r \rightarrow 0 \)

As \( r \) approaches zero, the term \( \frac{2V}{r} \) becomes infinitely large, while \( 2\pi r^2 \) approaches zero. Hence, \( S(r) \rightarrow \infty \).

Step 5: Conclusion on Behavior and Existence of Minimum

Since \( S(r) \rightarrow \infty \) as \( r \rightarrow \infty \) and as \( r \rightarrow 0 \), there must be at least one point in the interval where \( S(r) \) reaches a minimum value. We can find this by calculating the derivative and setting it to zero.

Step 6: Find Minimum Using Calculus

Calculate the derivative \( \frac{dS}{dr} \):\[ \frac{dS}{dr} = -\frac{2V}{r^2} + 4\pi r \] .Set this equal to zero and solve for \( r \): \[ 4\pi r = \frac{2V}{r^2} \]\[ 4\pi r^3 = 2V \]\[ r^3 = \frac{V}{2\pi} \]\[ r = \left(\frac{V}{2\pi}\right)^{1/3} \]Substitute \( V = 355 \) cm³ to find \( r \): \[ r = \left( \frac{355}{2\pi} \right)^{1/3} \].

Step 7: Calculate Optimal Radius

Computing numerically:\[ r \approx \left( \frac{355}{6.2832} \right)^{1/3} \]\[ r \approx 3.84 \text{ cm} \] .

Key Concepts

Calculus

Calculus is a powerful mathematical tool used to analyze change and find optimal solutions in various contexts like designing objects or systems efficiently. By utilizing derivatives and integrals, calculus allows us to understand how variables change in relation to one another, which is crucial for solving optimization problems like minimizing the surface area of a soda can.

In optimization scenarios, such as our soda can design problem, the derivative helps identify points where changes in variable values result in extremes (maximum or minimum). To minimize surface area, calculus involves setting the derivative of the surface area formula with respect to the radius to zero. This finds the radius value where surface area is minimized, provided that the behavior of the function meets specific criteria at that point, such as having the second derivative test confirm it's a minimum rather than a maximum or point of inflection.

Soda Can Design

Designing a soda can is not just about visual appeal but also about optimizing the use of materials.
The goal is to use the least amount of aluminum for cost efficiency while maintaining the structural integrity and volume requirements of the can.

By modeling the can as a cylinder, we can use standard geometric formulas to express volume and surface area:

• Volume formula: \(V = \pi r^2 h\)
• Surface area formula: \(S = 2 \pi r h + 2 \pi r^2\)

Substituting the constraint equation (volume) into the surface formula provides an expression dependent solely on the radius, enabling a calculus-based approach for optimization. In manufacturing, this balance between material cost minimization and operational parameters (like maintaining a fixed volume) is crucial for achieving optimal design.

Surface Area Minimization

Surface area minimization involves reducing the exposed area of the soda can to use less material and reduce costs.
The challenge is to achieve this while meeting the constraints of required volume. In our problem, the surface area of the can is a function of its radius, which changes how much material is needed.

To find the optimal surface area:

• We first express the surface area as a function of just the radius \(r\) by substituting the height from the volume constraint.
• Function to minimize: \(S(r) = \frac{2V}{r} + 2\pi r^2\)
• We calculate the derivative \(\frac{dS}{dr}\) to find critical points where the surface area is minimized.
• This critical point reflects the radius at which the can uses the least amount of material for a fixed volume.

Applying calculus here ensures that we are not just guessing but mathematically determining the most efficient design.

Constraint Handling

Handling constraints in optimization problems involves ensuring that all conditions or requirements of the problem are satisfied.
In the soda can problem, the primary constraint is that the can must hold a specific volume of soda, which is expressed as \(\pi r^2 h = V\).This is essential because any change in radius must be counterbalanced by a change in height to maintain this fixed volume.

Constraint handling ensures:

• Valid solutions: Only solutions that satisfy the constraints (such as fixed volume) are considered.
• Feasibility: Adapting the design (like changing the radius) upholds the core requirements of the can.
• Mathematical incorporation: Constraints directly influence the problem-solving approach by integrating into core equations, resulting in the solved height as \(h = \frac{V}{\pi r^2}\) before entering the calculus optimization stage.

In summary, constraint handling ties the theoretical math to practical requirements, ensuring that the optimized solution is feasible, functional, and efficient.