Optimizing the Design of a Cylindrical Can with Calculus

Calculus is a powerful mathematical tool that is essential for solving many real-world problems in various fields such as physics, engineering, economics, and biology. One specific example where calculus is crucial is in optimizing the design of a container to minimize material use while maximizing volume. This article will walk you through a practical example of designing a cylindrical can using calculus, highlighting the importance and application of this branch of mathematics.

The Problem: Designing a Cylindrical Can

A manufacturer wants to design a cylindrical can that holds a fixed volume of 500 cubic centimeters (cm3) while using the least amount of material for the can's surface area. This problem is a classic example of optimization, which is a fundamental concept in calculus.

Steps to Solve Using Calculus

1. Define the Variables

Let r be the radius of the base of the cylinder.

Let h be the height of the cylinder.

The volume V of the cylinder is given by the formula:

V pi r^2 h

2. Set the Volume Constraint

Since the volume is fixed at 500 cm3, we have:

pi r^2 h 500

From this, we can express h in terms of r:

h frac{500}{pi r^2}

3. Surface Area Formula

The surface area A of the cylinder, including the top and bottom, is given by:

A 2pi r^2 2pi rh

Substituting for h gives:

A 2pi r^2 2pi r left(frac{500}{pi r^2}right) 2pi r^2 frac{1000}{r}

4. Optimization

To find the minimum surface area, we need to take the derivative of A with respect to r and set it to zero:

frac{dA}{dr} 4pi r - frac{1000}{r^2}

Setting the derivative equal to zero:

4pi r - frac{1000}{r^2} 0

Rearranging gives:

4pi r^3 1000

Solving for r:

r^3 frac{1000}{4pi} quad Rightarrow quad r left(frac{250}{pi}right)^{1/3}

5. Second Derivative Test

To confirm that this critical point is a minimum, we can compute the second derivative:

frac{d^2A}{dr^2} 4pi frac{2000}{r^3}

Since both terms are positive, A has a local minimum at this value of r.

6. Final Dimensions

Once r is found, substitute back to find h using the volume constraint.

Conclusion

This problem illustrates how calculus is used to find optimal solutions under constraints, a common scenario in engineering and manufacturing. Without calculus, finding the minimum surface area for a fixed volume would be significantly more challenging if not impossible using algebra alone. This example demonstrates the practical application of calculus in solving real-world problems.