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Calculus Applied to Manufacturing Design

Suppose you are in the can manufacturing industry. Your job is to create cans according to the specifications of canned goods suppliers who will put their products in your cans and distribute them for sale. Now suppose one of those canned good suppliers comes to you and says, “We need a can that can contain a 300 mL volume of our product. Joe Shmoe says he can provide us these cans for $X. What can you do?” If you want the canned good suppliers business you will have to underbid Joe. (Provided you can’t sale them on some other facet of your company, i.e. service, quality, etc.) How will you do it? Where can you cut cost? What is the most inexpensive can you can make to satisfy the spec? In the real world this is a very complex question because it involves labor, facility, process, and material expenses along with several others. For the sake of this exercise however, we will use the principle of the derivative to optimize our can design in order to minimize material cost.

Now let’s assume you are set up to produce a standard cylindrical can with a top and a botttom. (I realize the canned goods supplier will have to add their product before the top is sealed and I realize that actual cans have a small lip on the top and bottom, but for the sake of simplicity…) To minimize the cost of the cans we want to minimize the amount of material used. But what shape cylinder will contain the required 300 mL but use minimal material? A short fat can or a tall thin can? How “tall” or how “short”?

First, realize the total material for the can can be determined by finding the area of the metal. For a cylinder this is the area of the bottom plus the area of the top plus the area of the side.

We can rewrite the area of the top and bottom as:

If you imagine removing the top and bottom, cutting the side along a vertical line, and laying it out flat you can see that the area of the side is the area of a rectangle with the length of one side equal to the height of the can and the length of the other side equal to the circumference of the top or bottom. So its area is:

Putting this all together, the total area of a can is given by:

Now the volume of the cylinder is given by:

If we rearrange this expression for the height, h we get:

Now we substitute this into our expression for the area to eliminate the h and make the whole thing in terms of 1 variable. (r – because the volume is given to us in the spec)

This equation tells us how much area a can has of a particular volume with any radius. With this in hand we can now use the derivative to optimize (maximize or minimize) the can design and find out which radius will give us the smallest area and thus the most inexpensive (as far as material costs go) cylindrical can that can contain the prescribed volume. Taking the derivative of the area equation using the power rule we have:

Now we set this equal to zero to find the minimum area. Solving for r we get:

Knowing that the volume is 300 mL our radius is:

Putting this back in our equation for h we get a height of:

These are the dimensions for our can of minimum material. Below I have tabulated and graphed some other values for the radius and associated heights that yield the necessary volume and show that the resulting surface area is larger than our optimized design.

One other beautiful thing about this technique is that I have a general formula for an optimized cylindrical can for any volume. I can just put in whatever the specification says and get my design.

Although this is greatly simplified, I hope this provides another demonstration of the real world applicability of these techniques.

Basic Derivative Rules

Previously we have discussed how a derivative is found using the limit, why the derivative is useful, and how to find it. Now to attempt to simplify or shortcut the derivative finding process, we will look at some general rules that can be used instead of going through the entire limit step-by-step every time.

The first derivative we will look at is the derivative of a linear function in slope-intercept form.

Where m is the slope and b is the y-intercept. Using the limit definition for the derivative we have:

Therefore the derivative of any linear function is, as we would expect, equal to the slope of that line. The next derivative we will look at is a special case of the linear function. If the slope is zero then the linear function is a constant function.

As above, the derivative of a linear function is its slope therefore the derivative of a constant function is also equal to the slope which is zero.

The next rule we can look at is a single independent variable raised to any power. Hence this rule is often called the Power Rule.

Applying the limit process we obtain:

Notice how we used the Binomial Theorem to expand the first term in the numerator. At this point I want us to notice a certain pattern in the coefficients that can be seen in Pascal’s Triangle. The first coefficient will always be 1 (c1=1). This can be seen by looking at the left edge of the triangle. Every first number is 1. Secondly if you look at the diagonal row parallel to, directly adjacent to, and to the right of this first row of 1’s, you notice that the numbers count sequentially as you descend the horizontal rows of the triangle: 1, 2,3,4,5,..etc. These numbers correspond to the second coefficient in the binomial expansion. If we label the horizontal row of the very first 1 at the top of Pascal’s Triangle zero and then incrementally as we go down the triangle, the second row will provide the coefficients for the binomial squared, the third row for the binomial cubed, the forth row for the binomial to the fourth power, and so on such that the nth row gives the coefficients for the binomial to the nth power and the second coefficient will be n also (c2=n). From here we have:

So you can see that the derivative of any variable to an exponent is equal to that same variable times the original exponent to the power of one less than the original exponent. For example:

In summary we can use the following rules to make our lives easier with regard to finding derivatives.

Next time we’ll look at some more helpful derivative rules.

Maximizing and Minimizing

In the last post about the derivative we emphasized its utility with some examples about how it can be used to find some relationships between position, velocity, and acceleration or to maximize profit and minimize cost in numerous different practical situations. Well now I want to discuss that in more depth and provide an example. Last time we used the following definition for the derivative (or slope) using the limit:

With this we can find the exact slope at any point on a curve. Now consider the following graph. Each of the red lines represents the slope at that point with a tangent line

Consider what the slope of such a tangent line would be at a minimum or maximum on a curve. That’s right, the line would be flat and horizontal like this:

Therefore the slope at these points is zero. With this geometrical understanding of a minimum or maximum we can find the derivative and then determine where the derivative (or slope) is equal to zero.

Now let’s consider an example function that relates the production level of a manufacturer to the potential revenue per unit. Obviously we want to maximize revenue so the key will be to find the maximum. Here is the function:

Now you might ask, “Well how in the world did we know this tells us about this relationship?” The short answer is that after some experimenting by the company at various different production levels and measuring the resultant revenue, the corresponding data points were plotted and the following function was mathematically obtained as a best-fit approximation to describe those data points. (More on this in a future post.) So assuming this is the correct function that we are working with, let’s first find the derivative.

Now this derivative gives us the slope at any point along the curve of the original function above. Where does this new function equal zero? This is the same as finding the roots of the equation which is a typical problem in algebra. Here is how it’s done based on our example.

Now if we put these two production level values into the original function it will tell us the revenue produced. Here are the results.

Therefore production level 2 provides the maximum revenue potential. (The root zero is an interesting point called an inflection point. But that’s another topic.) To demonstrate that this process has in fact located the maximum, here is a table of revenue values at other production levels and a graph of the function.

Next time we will look more in depth at the derivative and some basic rules or short-cuts we can use so that we don’t have to go through the entire limit process every time we want to find the derivative.

The Derivative and Why You Should Care

The first post in this series discussed the idea of the limit. And we tried to present it in a fairly simple, intuitive manner. We also discussed how a lot of the difficulty people ascribe to to calculus is due to the symbology. Above in the header you will see 4 different ways of writing the first derivative. However don’t be dismayed, just stick with me and we’ll see the derivative is not really that bad. Now with reference to the limit you might ask, what does getting really close to some point of anomaly have to do with derivatives? And what is a derivative anyway? And why should I care about all this math-ese? I suppose to motivate you to read on, I will first speak toward the usefulness of this concept.

The derivative allows us to consider the way one quantity or variable responds with regard to another quantity or variable. For instance, consider distance and speed. The more distance I cover in a period of time, the greater my speed. (For more on the relationship between distance, time, speed and acceleration see this post.) Or how about manufacturing: the less material I waste, the less my costs. Or maybe fuel economy: at a low speed I consume a lot of fuel. At a really high speed I consume a lot of fuel. But at some optimum intermediary speed, I maximize my fuel economy. Or maybe designing a structure for maximum strength with minimum weight. Or an electrical system that delivers adequate power with minimum infrastructure. All of these problems and many more utilize the concept of the derivative to arrive at their solutions.

Now with the hope that you are sufficiently convinced of its usefulness, let’s look at what exactly and derivative is. Again, I want to present it in such a way that you can intuitively grasp the idea. Take a look at the following graph.

Now the red line represents the function or the relationship between the input or independent variable, x, and the output or dependent variable, y. There are two of these key points that we want to look at. First at x=a, the function value is y=f(a). We use generic symbols so it will work for any function at any point. We will take a look at a particular function shortly to try and see this less abstractly, but for now let’s continue on. The second point is at x=a+h we get y=f(a+h). Now suppose we are curious to find the slope of the function curve (red line) at the point (a,f(a)) (the line labeled T). A rough approximation could be made by finding the slope of the line between the two points we mentioned. Recall that the slope of a line is equal to the rise divided by the run. The rise then will be equal to the difference in the two y-values, in particular f(a+h) – f(a). The run will be the difference in the x-values, in particular a+h-a which is equal to h. Therefore the approximate slope could be written as:

Now look at the graph and imagine that the point (a+h, f(a+h)) moved closer and closer and closer to the point (a,f(a)). Wouldn’t this give us a better approximation of the slope at (a, f(a))? I bet you have already noticed what concept I’m alluding to. That’s right, the limit. If we use the limit to bring the two points closer and closer together (in fact we will bring them so close together that the distance between them approaches nothing) we will have the exact slope at the first point. This is how we write that:

Now I promised an example so let’s look at a simple one. Let’s use:

Using the definition above we have:

So just like that we have found the slope or derivative of the function x squared which is 2x (“a” was just used above for a particular point, but it could really be any point “x”). In this same fashion we can find the derivative or slope of any function (provided it’s continuous – more on that later.)

But now you might say, “What does finding slope have to do with all the useful stuff you talked about at first?” We will dig into that next time.

The Limit: For When You Get Really Close, But Not Quite


Mention calculus and many people’s eyes glaze over and they fain their inability to handle such “difficult” and/or “boring” math. In actuality calculus is no more difficult than any elementary math that most people did very well at some point during their middle or high school educational experience and it is only boring because teacher typically fail to demonstrate the useful, relevant purpose for all the mathematical mumbo-jumbo. I think a lot of the initial hesitancy is also a result of unfamiliarity with the symbols that are used. I hope to explain in relatively simple terms some of the terminology and symbology associated with the study of calculus and show how its not that complicated and why it is useful. In this post we will start with the fundamental idea of a limit.

Starting with the basic idea of a function, f(x), where f is the function and x is the input variable or what is sometimes called the independent variable (Independent because it does not depend on anything else. We can put in the function whatever we want.) We often see a function written something like this:

Here is the graph or plot of this function:

It appears that the graph is a simple parabola however if we look at the function definition closely we see that an x-value of 1 will cause us a problem. We will get a zero in the denominator and that can’t be. Therefore this function is not defined at x=1. This is where the limit comes into play. It is very useful for looking at these odd points or areas and drawing some conclusions about what is going on there. Fundamentally what the limit does is says, “Well at this point there seems to be no value, but what about really close to that point…what about even closer…and closer…and closer. Basically what happens as we get super close to the input value without actually using the value.” We write this idea with the following symbols:

where f(x) is still the function “lim” is the limit operator and L is the limit value. The subtext x->a means the limit as the input x gets really close to the value a. Let’s use the above function as an example and consider its limit as it approaches the undefined point.

To get an intuitive idea of what we are doing let’s use some values that are really close to 1 and see what we get.

As you can see here it appears that as we get really close to 1 for our input value, the output or function value seems to approach 1. Therefore in a non-rigorous way we have an idea about what the limit means and what it is and we would write the limit for this as follows:

Also note here that in the table I approached the undefined input value from both above and below. If you are looking at the graph of the function this would be the same as saying I approached the undefined point from the left and from the right. This is important to note for future discussions about the limit. I hope you have seen from this post that these ideas are not all that scary and can be understood fairly easily. In future posts we will look at the limit in more detail, see how it is fundamental to all of calculus, and see how calculus is actually practical and useful.

Basic Kinematic Relationships

The study of kinematics is the study of the relationships between different elements of motion like position, velocity, and acceleration. Kinematics does not consider the forces that cause these motions, only the motions themselves. In this discussion I will use the symbol s for position, v for velocity, and a for acceleration. If a body travels from one position to another position and that trip starts at one time and ends at another time, we can write the average velocity for the body as:

However realize that this is the average velocity. The velocity at any point during the trip can vary because we are only considering the beginning and ending positions and times. To find the instantaneous velocity we use the same theory but we use calculus to express the change in position with respect to time using differential elements.

In much the same way we can express the average and instantaneous acceleration of a body as follows.

Using separation of variables on the last equation we can develop an integral relationship between acceleration, velocity, and time.

This says the ending velocity is equal to the beginning velocity plus the integral of the acceleration with respect to time. If the acceleration is constant through the entire time period we get:

Next let’s do the same separation and integration to the position and velocity differential equation.

Now we could proceed and find the relationship between position and constant velocity as we did above, but I think that step is obvious and I leave that to the reader. Here I want to make a substitution and come up with a different relationship. We will substitute the above equation relating velocity and constant acceleration into the last integral as follows.

This last equation can tell us the final position of a body if we know the original position, the original velocity, the constant acceleration rate, and the time of the trip.

Finally we will develop one more kinematic formula. We start again with the differential equation relating velocity and position. Then we will use a chain rule and variable separation with integration to reach our destination.


Notice that this equation does not require knowledge about the time. All we need is the original velocity, the constant acceleration rate, and the change in the position to find the final velocity.

I hope you see from this discussion that kinematics is a very simple, fundamental exercise in mechanics. If we boil all of this down, we can solve any problem in general by simply applying the principles of calculus to the  two fundamental differential equations:

to formulate whatever relationship we need.