Pontos de extremo e otimização

Optimization is the process of extremizing continuous functions. Assume the following sample function.

$f(x) = 4x^2 - x^4$

We want to find the maximum of this.

The brute-force approach to this is to evaluate the function at every possible point and find the lowest number, but there are infinitely many points, which would make this task hard. We need to find some candidate points.

A continuous function will extremize when it has become so extremized it stops changing, whenever its derivative is equal to $0$.

$\frac{d}{dx} (4x^2 - x^4) = 0$

$8x - 4x^3 = 0$

$x = \{-\sqrt{2}, 0, +\sqrt{2}\}$

This tells us that we have extreme values at $x = \pm\sqrt{2}$ and $x = 0$, but we need to determine whether these are local maxima or minima.

If a point is a local maximum, around that point the function will go up and then go down. If it is a local minimum, the function goes down and then goes up (look at the graph). What we now need to do is figure out where the derivative is positive and negative.

$(-\infty, -\sqrt{2} ) \hspace{10mm}+$

$(-\sqrt{2}, 0 ) \hspace{16mm}-$

$(0, +\sqrt{2} ) \hspace{16mm}+$

$(+\sqrt{2}, +\infty ) \hspace{10mm}-$

This makes the two square root... roots local maxima, and the zero root a local minimum.

The function goes go $-\infty$ from both sides, which means that there should be an absolute maximum. Out of all of the local maxima that exist, the largest of those is the absolute maximum.

The absolute maximum of $f(x)$ is $f(\pm\sqrt{2}) = 4$.

We have 100 blocks of wood to build a rectangular... fence. What is the maximum area we can get?

We can get two equations out of this: a constraint and an area. Assume $x$ and $y$ are side lengths, and we want to extremize the area $A$.

$A = xy$

$100 = 2x + 2y$

We can handily express the second equation in terms of $y$, so we can get the first one in terms of only $x$.

$y = 50 - x$

$A = 50x - x^2$

Now we have to find the derivative, find candidate points, determine whether they're maxima or minima, and finally find the absolute maximum area possible.