Studying the existence of minimizers

[malekhammou]

Hello Everyone
In one of my assignments, I am asked to study the existence of possible minimizers and finding possible ones for an optimization problem represented by a function f (Find attached the statement of the exercise). I proceeded by finding the points where the gradient is equal to zero and where the Hessian is definite positive.
I am wondering if you could possibly provide help and guidance.

 

[HallsofIvy]

What does "minimizer"mean? Is that "values of x and y that make those expressions minimal? For the first one it should be obvious that, since a and b are positive, any non-zero values of x and y make the expression positive while (0, 0) makes it 0.

For the second, you say "I proceeded by finding the points where the gradient is equal to zero and where the Hessian is definite positive."
Okay, that sounds good! What did you get?

 

[malekhammou]

Yes, the first function was obvious. For the second, I calculated the gradient and applied the first-order condition where Gradient(x,y)=0. It resulted in solving a system that yielded (0,0) as values of x and y. Applying the second-order condition where <Hess(0,0)(h1,h2),(h1,h2)> is positive for any (h1,h2) belonging to R^2 resulted in the following : 2(h1-h2)^2-2h1*h2

 

[HallsofIvy]

I don't know what your "\(\displaystyle h_1\)" and \(\displaystyle h_2\)" are.

Yes, setting the gradient gives (0, 0) as the only critical point.

The Hessian is \(\displaystyle \left|\begin{array}{cc} \frac{\partial^2 f}{\partial x^2} & \frac{\partial f}{\partial x\partial y} \\ \frac{\partial^2 f}{\partial x\partial y} & \frac{\partial^2 f}{\partial y^2}\end{array}\right|= \left|\begin{array}{cc} 2 & -3 \\ -3 & 2\end{array}\right|= 4- 9= -5\) so (0, 0) is a "saddle point".​

​

You could also do this by "completing the square". The coefficient of x is "-3y". Half of that is -3y/2 and the square is \(\displaystyle 9y^2/4\).​

\(\displaystyle x^2- 3xy+ y^2= x^2- 3xy+ \frac{9y^2}{4}- \frac{9y^2}{4}+ y^2= \left(x- \frac{3y}{2}\right)^2- \frac{5}{4}y^2\)

That is a hyperbolic surface with critical point where x- 3y/2= 0 and y= 0, (0, 0). It is increasing in the direction x= 3y/2 and decreasing on the y= 0, the x-axis.

 

[malekhammou]

Thank you very much! Your explanation is very clear and straightforward. Apologies for the ambiguity regarding "h1" and "h2", I was referring to a general case and I made a mistake.

Again, thank you for your time and your help.
Regards,
Malek