Optimization with Derivatives

# Optimization with Derivatives

Optimization problems often deal with the question, "what is the largest/greatest (or smallest/least) given some constraint", in some manner that a function representing a problem can take. We've already looked at Maxima and Minima, so solving these problems shouldn't be too much of a hassle. Of course, we have a strategy for solving optimization problems.

• 1. First read through the problem carefully and identify any important relevant information.
• 2. Find any constraints in the problem, that is, a value in the problem that must always remain true.
• 3. Draw a diagram to illustrate the problem and give you a visual.
• 4. Find the maximum/minimum.

## Example 1

Joe wants to build a rectangular fence attached to his house. Joe has $400m$ of material, and wants the area enclosed in the fence to be a maximum. Find the dimensions of the field enclosed by the fence.

We first identify that we are looking for the maximum area enclosed by the fence, and that we have precisely $400m$ of material (our constraint) to work with. Let $x$ be the two sides of the fence attached to Joe's house, and let $y$ be the other side of the fence. The diagram below illustrates the problem:

We therefore get that we want to maximize the area $A = xy$, while we have a constraint that $2x + y = 400$. Now let's rewrite our constrait to be $y = 400 - 2x$, and thus:

(1)
\begin{align} A = xy \\ A = x(400 - 2x) \\ A = 400x - 2x^2 \\ A' = 400 - 4x \end{align}

If we let $A' = 0 = 400 - 4x$, we get that $x = 100$. Therefore, when $x = 100$, $y = 200$, and the maximum area of the field will be $20000m^2$.