Min, max and saddle points of a function

Question

I want to find all local minima, maxima and saddle points of the function $f(x,y)=(x-y)(1-xy)$

Therefore I wanted to set the partial derivatives to zero to get all possible points.

First I calculated the derivatives $f_x(x,y) = y^2-2xy+1$ and $f_y(x, y) = -x^2+2xy-1$

From $f_x$ I got $x=\frac{y^2+1}{2y}$ and I used it in $f_y$ to be able to determine the values of $y$.

$-\frac{y^2+1}{2y}^2+2y\frac{y^2+1}{2y}-1=0$

Then I got $y^4-\frac{10}{3}y^2-\frac{1}{3}=0$

I substituted $z = y^2$ and got $z^2-\frac{10}{3}z-\frac{1}{3}=0$

When I wanted to solve this, it leads to $z_1= \frac{5+\sqrt(28)}{3}$ and $z_2= \frac{5-\sqrt(28)}{3}$. As I am not allowed to use a calculator, I think that I've done something wrong as it's for example not very easy for me to determine the result of $\sqrt(28)$ without using one, but I'm not able to find what I may have done wrong.

Answer 1

I finally found the error in a sign. When correcting it, it leads to

$y^4-\frac{2}{3}y^2-\frac{1}{3}=0$

which leads to the values $y_1 = 1$ and $y_2 = -1$.

Then the possible points are $(1,1)$ and $(-1,-1)$.

The point $(1,1)$ is a minimum, $(-1,-1)$ is a maximum and there are no saddle points.

Answer 2

You are ok expressing it as a square root...no more is needed. By the way, using taylor series is a great way of approximating functions. Or using differentials (easier):

Call $f(x)=\sqrt{x}, \space x=9 ,\space dx=1 \space:$ $$f(x+dx)\approx f(x)+f'(x)dx$$ $$\sqrt{10}\approx\sqrt{9}+\frac {1}{2\sqrt{9}}*1$$ $$\sqrt{10} \approx \frac {19}{6} \approx 3,16$$

If you use the binomial expansion (taylor) you can get as close as you want. But in math problems, it is ok to leave answers expressed in square roots, exponentials, or whatever function you can plug in a calculator later.