Max and min of $f(x,y)$ defined in $\mathbb{R}^2$ I consider the function $f(x,y)=3x^2+5y^2+xy+3x-y+1$.
I have found a min in $(-31/59,9/59)$. How can I say is a global min?
 A: As $f$ is a quadratic function, the easiest way to find everything you want is a Gauss canonization :
\begin{align}
  f(x,y) &= 3\left(x^2+\frac{xy}{3}+x\right) + 5y^2-y+1 \\
         &= 3\left(x+\frac y6+\frac12\right)^2 -3\left(\frac y6+\frac12\right)^2+ 5y^2-y+1 \\
         &= 3\left(x+\frac y6+\frac12\right)^2 +\frac{59}{12}\left(y^2-\frac{18}{59}y\right) +\frac14\\
         &= 3\left(x+\frac y6+\frac12\right)^2 +\frac{59}{12}\left(y-\frac{9}{59}\right)^2-\left(\frac{9}{59}\right)^2+\frac14
\end{align} 
This proves :


*

*that the function $f$ has a unique minimum when $y=\frac9{59}$ and $x+\frac y6+\frac12=0$, so $x=-\frac{31}{59}$,

*that this minimum is $\frac14-\left(\frac{9}{59}\right)^2=\frac{3157}{13924}$.
I hope my computations are correct, but you can see the general idea.
A: Hint:
You can decompose this quadratic function as a sum of squares of linear functions with positive coefficients by Gauß' method. 
A: This method works for identifying all extrema for a function. First find the critical points by calculating $$\begin{align*} f_x &= 6x + y + 3\\ f_y &= 10y + x - 1\end{align*}$$
We need to find the critical values, these are found where $f_x = f_y = 0$. So we can set the equations equal to each other and solve for the function
$$\begin{align*} 10y + x - 1 &= 6x + y + 3 \\ 9y &= 5x + 4\end{align*}$$
Now every point on this line is a critical point. To find the global minimums we must find the $(x_0, y_0)$ such that $$f_{xx}f_{yy} - f_{xy}f_{xy} > 0$$
when it is evaluated at $(x_0, y_0)$. Now $$\begin{align}f_{xx} &= 6 \\ f_{yy} &= 10 \\ f_{xy} &= 1\end{align}$$
So we have $f_{xx}f_{yy} - f{xy}f_{xy} = 6(10) - 1 = 59 > 0$. Hence every critical value on $9y = 5x + 4$ is a global minimum because there are no points off this line that are smaller than it. 
Lastly we show that $(-\frac{31}{59},\frac{9}{59})$, is on the line. Just verify that $$9\Big(\frac{9}{59}\Big) = 5\Big(-\frac{31}{59}\Big) + 4$$
which is true.
A: You have $f(0,0)=1$ and $f(x,y)\geq 10$ when $x^2+y^2\geq R^2$ for  sufficiently large $R$. Fix such an $R$.  The function $f$ assumes a global minimum on the compact set $B_R:=\{(x,y)|x^2+y^2\leq R\}$, and this minimum value is $\leq1$, hence is taken at an interior point  $p\in B_R$. It follows that $p$ is a critical point of $f$, and as you have found just one critical point this has to be the point where the global minimum of $f$ on $B_R$, hence on ${\mathbb R}^2$, is taken.
