Calculus is usually the first time students are introduced to the field of optimization. Optimization problems are arguably the hardest aspect of undergraduate introductory calculus courses because they require you to extract information and relationships among variables given a set of initial conditions or a prompt. This makes it really hard because very few instructors take a methodical approach. Here are my recommended steps for solving optimization problems:

*Look at the equation to find the critical value/variable**Look at the parameters/initial conditions/ whatever you are given and extract information and unknown relationships.**Set up an equation that expresses your parameters as a single variable (critical value)**Derive for the max/min.*

Lets try a sample problem where a piece of paper is folded (x) in such a manner that it creates a crevice (L).. ( only a mathematician would think of this stuff…)

Our ultimate goal: **express L in terms of X**

There are three main relationships we can deduce from the picture given, ill call them triangles A B and C.

Since we need to define L in terms of X we have to first look at the expression for L, in triangle B

In order to express L in terms of X we need to find a way to express Z in terms of X, thankfully euclidean geometry uses linear scaling and proportionality, so we can observe that both Z and X are hypotenuses, therefore we can say they are equivalent, and from there using the heights given, express Z in terms of X., which we can use to express L in terms of X.

Next we want to clean up our expression to make the derivation of L as easy as possible.

now that we have defined L in terms of X all we have to do now is derive and solve for the absolute minimum value in this problem since it asks to have L as SMALL as possible.

Using the chain rule and the quotient rule, we can see that L = undefined when x = 0, 3/4ths of K and 1/2 of K.

Since X being zero would be impossible for the purposes of this problem given the prompt we can omit it and proceed using the same procedure for solving any optimization problem, namely finding out the values of L’ over intervals of the two remaining critical values.

In this example we tested the values for x using x = 2/3rds of K and x = K

Since the value of L’ goes from negative to positive X = 3/4ths of K is an absolute minimum and therefore the value we need to make L as small as possible.