Max and min of $f(x,y)$ defined in $\mathbb{R}^2$

Question

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?

Answer 1

Hint:

You can decompose this quadratic function as a sum of squares of linear functions with positive coefficients by Gauß' method.

Answer 2

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 :

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

2. 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.

Answer 3

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.

Answer 4

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.