Minimum value of $\displaystyle f(x,y)=2x^2-3xy+5y^2-2y+6$

Let $\displaystyle k=2x^2-3xy+5y^2-2y+6$

$\displaystyle 2x^2-3xy+5y^2-2y+6-k=0$

For real values of $x,$ equation has real roots $D\geq 0$

$\displaystyle 9y^2-4(2)(5y^2-2y+6-k)\geq 0$

$\displaystyle -31y^2+16y-48+8k\geq 0$

$\displaystyle 31y^2-16y+48-8k\leq 0$

I did not know how to proceed further, Help required.


2 Answers 2


We have that by completing the squares

$$2x^2-3xy+5y^2-2y+6 =2\left( x-\frac{3}{4}y\right)^2+\frac{31}8\left(y-\frac8{31}\right)^2+\frac{178}{31}\ge \frac{178}{31}$$

with equality at

$$x-\frac{3}{4}y=0\;\land\;y-\frac8{31}=0 \iff (x,y)=\left(\frac6{31},\frac8{31}\right)$$

  • 1
    $\begingroup$ To be fully rigorous you should point out that this value $178/31$ is in fact achieved at some point i.e. $(6/31 , 8/31)$ $\endgroup$ Commented May 17 at 17:42
  • $\begingroup$ @RupertRybka Yes indeed, it seems clear from the expression but I can add that! Thanks $\endgroup$
    – user
    Commented May 17 at 17:43

We can find partial derivative of $f(x,y)$ with respect to $x$ and $y$. $$\frac{\partial f}{\partial x}=4x-3y$$ $$\frac{\partial f}{\partial y}=10y-3x-2$$

This will give us two linear equations and the intersection points of those two linear equations are the required values of $x$ and $y$ for which the ljnear equations disappear.

Taking the second partial derivatives, $$\frac{\partial^2 f}{\partial x^2}=4,\frac{\partial^2 f}{\partial x\partial y}=-3,\frac{\partial^2f}{\partial y\partial x}=-3,\frac{\partial^2f}{\partial y^2}=10$$

So the Hessian matrix is $$\begin{pmatrix} 4 & -3 \\ -3 & 10 \end{pmatrix}$$

It's determinant is $40-9=31>0$. So the Hessian is a positive definite matrix by Sylvester’s criterion and the point where the partial derivatives vanish is a point of global minimum.

So $$y=\frac{8}{31},x=\frac{6}{31}$$ Putting these values in we get the minimum value to be $$2\left(\frac{36}{961}\right) -3\left(\frac{6}{31}\right)\left(\frac{8}{31}\right)+5\left(\frac{64}{961}\right)-2\left(\frac{8}{31}\right)+6.$$ This equates to $$\frac{178}{31}.$$

  • $\begingroup$ You've found a turning point in both directions. Why must it be a global minimum? Why can't it be a local minimum, saddle point, or a global maximum? $\endgroup$
    – Calvin Lin
    Commented May 17 at 19:20
  • $\begingroup$ It has to be a global minimum as when you find the second partial derivative wrt x and y, they both are positive, so it has to be the global minimum. $\endgroup$ Commented May 18 at 9:32
  • $\begingroup$ Can you show that explicitly in your writeup? Otherwise the solution is incomplete. $\quad$ Furthermore, note that all that shows is that we have a local minimum. EG $- x^2(x^2-1)$ at $ x= 0 $ has a first derivative of 0 and a positive second derivative, but it's not a global minimum. $\endgroup$
    – Calvin Lin
    Commented May 18 at 17:05
  • $\begingroup$ I will add an mathematical proof for it being a global maximum $\endgroup$ Commented May 18 at 17:22

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