Optimize Your Savings: Unraveling the Math Behind Cost-Efficient Cylindrical Can Design Using Calculus

Introduction: Welcome to this intriguing exploration of calculus and its real-world applications! Before we dive into the fascinating world of math, I’d like to clarify that I’m not a math wizard myself. In fact, the following calculus problem and its solution have been generated by the impressive GPT-4 AI. That said, I encourage any math aficionados among you to verify or debunk the answer provided. So, without further ado, let’s delve into the mathematical mystery and see what GPT-4 has in store for us!

Here’s a challenging calculus problem involving optimization, which is a common real-world application of calculus. Let’s consider the problem of designing a cylindrical can to minimize the production cost.

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Exploring a Complex Calculus Problem and Its Real-World Application

Problem: A cylindrical can with a volume of 500 cubic centimeters (cc) is to be designed. The cost of the material for the top and bottom is $0.05 per square centimeter, while the cost of the material for the sides is $0.02 per square centimeter. Determine the dimensions of the can that will minimize the total cost of producing it.

Setting Up the Calculus Problem

Let’s set up the problem with the following variables:

  • r: radius of the base of the cylinder
  • h: height of the cylinder
  • V: volume of the cylinder = 500 cc

Formulating the Required Equations

Now, let’s write out the equations we’ll need.

  1. Volume equation: V = πr^2h
  2. Surface area equation: S = 2πrh + 2πr^2 (area of the sides + area of the top and bottom)
  3. Cost equation: C = 0.02(2πrh) + 0.05(2πr^2)
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Eliminating a Variable Using the Volume Equation

First, we want to eliminate one variable. Since we know the volume, we can solve the volume equation for h:

h = V / (πr^2) = 500 / (πr^2)

Now, substitute this expression for h in the cost equation:

C(r) = 0.02(2πr(500 / (πr^2))) + 0.05(2πr^2)

Simplify the equation:

C(r) = (20 / r) + 0.1πr^2

Minimizing the Cost Equation

Now, we need to minimize the cost equation. To do this, we’ll find the critical points by taking the derivative of the cost function with respect to r and setting it equal to zero:

dC/dr = -20/r^2 + 0.2πr

To solve for r, set dC/dr = 0:

0 = -20/r^2 + 0.2πr

Now, multiply by r^2 to clear the fraction:

0 = -20 + 0.2πr^3

Solve for r^3:

r^3 = 100 / π

Now, find r:

r ≈ 2.88 cm

Now that we have the optimal radius, we can find the height using the equation for h we derived earlier:

h = 500 / (π(2.88)^2) ≈ 5.77 cm

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Optimal Dimensions for Cost Minimization

So, the dimensions that minimize the cost of producing the can are approximately a radius of 2.88 cm and a height of 5.77 cm.

This optimization problem is just one example of how calculus can be applied to real-world scenarios, such as minimizing the cost of producing an object with specific requirements.