Summary of Optimization on Closed Intervals

Optimization is determining minima and maxima of a function. This is one of the many useful practical applications of calculus.

Given a function $f$ on a closed interval $[a,b]$ that is differentiable on the open interval $(a,b)$, the Extreme Value Theorem states that there exist values $m$ and $n$ in $[a,b]$ such that $f(m) \ge f(x) \ge f(n)$ for all $x\in [a,b]$. We call the value $f(m)$ the maximum value of of $f$ on $[a,b]$ and $f(n)$ the minimum value of $f$ on $[a,b]$.

The values of $m$ and $n$ must be one of the following:

Case 1: a boundary value, i.e. $a$ or $b$

Case 2: a value $x$ in the interval but not a boundary such that $f'(x) = 0$

Case 3: a value $x$ in the interval but not a boundary such that $f'(x)$ is undefined (this is not a concern if $f$ is differentiable.

So, to determine the minimum and maximum values of a function $f$, we do the following:

1. Take the derivative of $f$.

2. Solve the derivative on the interval $[a,b]$ and see if it is $0$ at any value in the interval. Use algebra to identify zeros of $f'$.

3. $a$, $b$, and any values found in Step 2 are what we call "critical values." Identify all critical values.

4. Find the value of $f$ at every critical value.

5. The minimum of the values found in Step 4 is the global minimum of $f$ on $[a,b]$. The maximum found in Step 4 is the global maximum of $f$ on $[a,b]$.

Example: (disclaimer: this does not represent real-world finance) Suppose that I have 1000 dollars and I have two options for my money:

Option 1: A future contract which will return 1.2 times the amount of money I put into it.

Option 2: A friend who is willing to give you $20\sqrt{x}$ if you pay him $x$ dollars. (Your friend loves calculus and is doing this almost solely for the purpose of teaching you calculus).

I want to maximize my money received at the end, assuming all my money must go into either Option 1 or Option 2.

Solution: Let $x$ be the amount of money put into Option 2. Then $f(x) = 20 \sqrt{x} + 1.2 \cdot (1000-x)$ is the money I receive. I want to find the maxima of $f$ on $[0,1000]$. The derivative is $f'(x) = \frac{10}{\sqrt{x}} - 1.2$ This has a zero at $\frac{625}{9} \approx 69.444$.

So, our critical values are 0, $\frac{625}{9}$ and 1000. Evaluating $f$ at each value, we find $f(0) = 1200$, $f(\frac{625}{9}) \approx 1283.33$, and $f(1000) \approx 632.46$. So, since $f(\frac{625}{9})$ is the largest of those, that means 1283.33 is the maximum amount of money I can have at the end, and it is achieved when I invest $\frac{625}{9}$ into Option 2 and $1000-\frac{625}{9}$ into Option 1.

This is only one way this technique can be applied!

(Video) The Closed Interval Method

by PatrickJMT

Here's a video by patrickJMT explaining the Closed Interval Method and how to use it to find absolute maximums and minimums.

He states that to find the absolute maximum and minimum values of a continuous function $f$ on the closed interval $[a,b]$:

1. Find the values of $f$ at critical numbers of $f$ in $(a,b)$
2. Find the values of $f$ at the endpoints of the interval
3. The largest is the absolute max; the smallest is the absolute minimum

The problem he looks at is:

Find the absolute maximum and absolute minimum of $f$ on the given interval: $$f(x)=\frac{x^2-4}{x^2+4};\ [-4,4]$$

First let's find the critical points. $$\begin{align} f(x)&=\frac{x^2-4}{x^2+4}\\ f'(x)&=\frac{\color{red}{[x^2-4]'}(x^2+4)-(x^2-4)\color{red}{[x^2+4]'}}{[x^2+4]^2}\\ f'(x)&=\frac{2x(x^2+4)-(x^2-4)2x}{(x^2+4)^2}\\ f'(x)&=\frac{2x^3+8x-2x^3+8x}{(x^2+4)^2}\\ f'(x)&=\frac{16x}{(x^2+4)^2}\\ \end{align}$$

Now look at where the derivative equals zero and is undefined. $$\begin{align} \frac{16x}{(x^2+4)^2}&=0\\ 16x&=0\\ x&=0 \end{align}$$ $$\begin{align} (x^2+4)^2&=0\\ x^2+4&=0\\ x^2&=-4\\ \text{No real }&\text{solution.} \end{align}$$

So the only critical point we have is $x=0$. We also have the end points to test $x=-4$ and $x=4$. Our next step is to plug all three of these points into our function and see what values we get out. $$\begin{align} f(-4)&=\frac{\color{red}{(-4)}^2-4}{\color{red}{(-4)}^2+4}\\ &=\frac35\\ f(0)&=\frac{\color{red}{(0)}^2-4}{\color{red}{(0)}^2+4}\\ &=-1\\ f(4)&=\frac{\color{red}{(4)}^2-4}{\color{red}{(4)}^2+4}\\ &=\frac35 \end{align}$$

So we have an absolute minimum of $-1$ and an absolute maximum of $\frac35$. It's okay that we get an absolute maximum at both of our endpoints.