This topic is not too hard to understand but can give students troubles as the algebra can get a bit intimidating.

The general idea in optimization is not to just find A solution, but to find the best solution. This is done by mathematically modelling a situation and then looking for a maximum or minimum in that model. While we will only tackle "simple" scenarios this same idea is used by engineers, economists, businessman, you name it anyone who wants or needs to know the best solution. The ideas and concepts can be generalized to more complex scenarios and frankly is one of the more truly useful ideas you'll encounter in Math SL.

To find the optimal solution we need to follow the 4 general steps below.

  1. Write an equation for what you are trying to optimize
  2. If the equation from #1 has more than one variable, find an equation of constraint or in other words find an additional equation that relates the two variables.
  3. Use the constraint to algebraically reduce the number of variables.
  4. Find the max or min as appropriate.

Classic Example:

A farmer has 800 m of fence and wants to create the largest rectangular enclosed area possible. The farmer is also going to use the edge of a river as one side of the enclosed area. This way the farmer only needs to use fence on 3 sides of the enclosed area. What is the maximum area the farmer can enclose?


Step 1
The area of the is what we are trying to maximize. We can write an equation for the area:

\begin{equation} Area = xy \end{equation}

Step 2
We also know that the farmer is "constrained" by only having 800 m of fencing. Which we can express in an equation as:

\begin{equation} 2y+x=800 \end{equation}

Step 3
This equation can be rearranged as $x=800-2y$ and then substituted into the area equation:

\begin{equation} Area = (800-2y)y = 800y - 2y^2 \end{equation}

Step 4
Now we need to find the maximum of the area function. This can be done by graphing and then finding the maximum OR by finding the first derivative and setting it equal to zero.

The derivative of the area is given by

\begin{equation} A' = 800-4y \end{equation}

Setting this equal to zero and solving for y results in $y = 200 m$ with $x = 400 m$.

More Examples

These will come or not. I'm feeling lazy. If folks have specific questions post them in the comments and maybe those can become examples.

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