We discuss how to model and then optimize that model, for real world style problems.
Video Lecture
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One of the useful applications of calculus is in the very broad area of optimization. But saying optimization is “an” application of calculus is kind of like saying the internet access is “an” important aspect of taking online classes. Optimization is an entire area of study in graduate level mathematics, and what we even mean by “optimization” can vary massively. As such, there isn’t a set formula or even sequence of steps that will always work. Nevertheless in this video we will aim to give a sequence of steps that will provide a general framework on how to approach some common types of optimization problems.
By its nature, optimization is a question about trying to get some “best result” under some kind of constraint.
For example, let’s say you work for a company that sells canned soup. Your company sells soup in a number of volumes, but your job is to determine how to design the containers. You want to see if it is best to use a cylinder like the traditional soup can, or a box like container or some other style. How do you figure this out? How do you figure out the best dimensions of each of these types that costs the least amount in materials? This is a classic optimization problem. You are trying to get the “best result”, finding the least cost, under the constraint that there is some specific volume you need to contain.
The problem with tackling optimization as a single topic though, is that it spans everything from maximizing volume with minimizing costs like our soup can, to finding a fastest path to get somewhere, to maximizing profit, to balancing fuel efficiency with speed when planning a flight. The applications are so diverse that our algorithm for approaching optimization problems will necessarily be vague, and not all steps or parts of steps may apply to every problem. All of this is to say, these are ... “more like guidelines than actual rules.”
With that caveat, let’s start.
- Step 1:
- The first step, which is the first step for almost any kind of math problem, is to try and sketch a picture of what you
are dealing with if possible. Whether this is a literal sketch of the situation, like a sketch of your soup can, or a sketch of
the general information and how they are related, like a sketch that is just a sequence of words with arrows that direct
which thing leads to which. The reason this is so helpful is that it helps you crystallize exactly what is involved in the
problem; what information is involved and roughly how one thing might related to another. This may seem silly, but this
is a great way to get a birds-eye view of the structure of the situation.
For our soup can example, we would sketch a cylindrical can, maybe with a nice little soup logo on the side!
- Step 2:
- The second step is to apply the language of mathematics to the problem. Start assigning variables to anything that
can be quantified, even if it isn’t something you may want to actually use. It’s best to overdo it at this step, and then cut
back later, because it is harder to figure out what is missing later than to realize when you don’t need to use something
you initially had written down.
For our soup can example, we would put down some dimensions, like height, and radius of the cylinder.
- Step 3:
- Step three: This is one of the hardest steps - you need to relate the key ideas into a function. You want to come up
with a function that represents the information you are trying to optimize. At this step, you don’t want to worry about
what is making the optimizing difficult (aka the constraint). For this step, you want a function whose output is the thing
you are trying to optimize. This is often referred to as the “Primary Function” or the “Optimizable Function”
In our soup can example, this would be the formula that tells you how much it costs to make a soup can with some given dimensions.
- Step 4:
- Step four is when we try and determine the constraint. This is whatever is stopping you from just going crazy and
getting some unreasonably extreme answer for your primary function.
In our soup can example, the reason we can’t have a cost of zero (by having a can with zero in every dimension) is because we need the soup can to hold a specific volume of soup. This required volume of soup is the constraint for the problem.
- Step 5:
- Step five is easily stated and often difficult to execute. Simply put, step five is to “use calculus to find the desired absolute extrema”. Often this is done by using the constraint relationship to reduce the number of different variables in your primary function - allowing you to turn it into a function of one variable. Once this is accomplished, you can use your techniques for finding absolute extrema to determine what the “best” solution is for that primary function (and where it occurs). Keep in mind though that often things like domain and especially domain restrictions, are very important here and may not be obvious in the equations themselves, rather they occur due to natural constraints - like how we can’t have a negative length, or negative volume for our soup can.
So, let’s consider our soup can example from start to finish.
- Step one: draw a picture
- We make a picture, like a wire diagram of a soup can.
- Step two: apply mathematical language
- In our case, we would add dimension lines for height and radius of the top, with variables “” and “” respectively.
- Step three: Build a primary function
- We want to optimize the cost, so our function should look like “Cost = (something)”. So, the natural place to start is to write out a formula for the cost to make our sketched can. Since the can includes a top and bottom, as well as the side, we need a formula for the surface area of the cylinder, which is: . Since the material is all the same cost per square inch, then that means our cost for the can is .
- Step four: Find the constraint
- The constraint is that the soup can must hold 32 cubic inches. So we should represent this by the volume formula, .
- Step five: Use calculus to find the absolute extrema
- As mentioned, this usually occurs by using the constraint to reduce
the original equation down to a single variable. In this case, we will solve for in our constraint equality and then plug that
into our primary function to get a function entirely in terms of .
Solve for in our volume formula: . Then substitute what we got into the in the cost formula to get...
Now that we have a function for the cost, which is only dependent on one variable (in this case, the radius) we can take a derivative and find our extrema. Note here though that there is an obvious domain restriction with , but this isn’t really an issue because we should be wary of trying to have a can with a radius of zero... that won’t work out too well for us. Additionally, there is also a restriction that can’t be negative, since it represents a distance. Thus we can conclude that .
Next we need to take the derivative to find the extrema. Taking a derivative then, we get:
Since we want to find the extrema for this function, which occur when the derivative is zero, we need to find the zeros of the function. To this end, we need to combine them into one fraction, which gives us:
So our zero occurs when , i.e. at . Using this, we can get from our earlier formula (from the constraint equation we isolated to get ) to get
So we’ve seen the general guidelines on how to solve optimization problems, along with an example. The key thing here though is that these guidelines are a good framework to solve an optimization problem, but every problem is bound to be a little different. Don’t memorize the algorithm and expect to follow it blindly, because inevitably whenever you try to do an optimization problem it will be at least a little bit different from every other optimization problem you’ve tried. For this reason, you should also make sure to do a lot of practice, and watch any example videos you can to get a better feel for the various ways these problems can be formulated, modeled, and manipulated.