Monthly Archives: September 2011
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.
Recently, I had the honor of being interviewed by Nayema Chowdhury at Engineering Because. They have done numerous interviews with engineers and engineering students from all around the world. They also have an online community as part of their website that allows you to meet some of these people. You can check out the site and read my interview here.