So we need to answer the question, whether to find local extrema, we have found two points of local maximum, two points of local minimum whether global maxima minima at the same time? This is not true here. To clarify the issue, we need to think about the behavior of this function z at infinity. So if we draw the coordinate plane and choose the direction of travel. So for instance, I can choosing the bisector straight line which passes through the origin at 45 degrees angle. So if for instance, I am moving along with this bisector and go to the infinity, then z x,x becomes x squared ln x squared, it goes to positive infinity. That means that this function has no global maximum because it never takes the greatest value. At the same time, if I choose another bisector and this time we take z at x minus x, the the function takes the negative value. When x moves to negative infinity, the function itself goes to negative infinity. That means the global minimum also doesn't exist. Let's consider another problem. This time the function z is like this. We need to find its global extrema if they exist. We follow the same route. We start with identifying the critical points firstly. In order to solve the system, let me solve for y in terms of x. So then it becomes and I will substitute into the second equation. So that will be a resulting equation having only one x, no y. So I'm simply rewriting substituting in this term the y-value. So I've y being placed here. Now we need to collect all the pose of number two. I have two cubed and this is two squared. Two squared makes two to the sixth x nine power. So well, clearly I get several points. The first point being A simply zero, zero. It was clear from the start that zero, zero for xy is a solution, one of the solutions of this system. After that I can cross out x in the left-hand side and the right-hand side, then I get equation x to the eighth equals one over two to the ninth power and that gives me two solutions, one positive another one negative. For positive, I have one over two to the power of the ninth-eighth. For negative, the same in absolute value but I have a minus in front. So these are my values of x. Need also to find y. So if I choose that, the y will be positive. So I have two squared and I raise this to the third power. Then I have something like 27 over 8. I need to subtract two. Twenty seven-eighths minus two makes one over eleven-eighths. A similar y-value for another one also but I teach minus sign. So I want to do it right here. So we have three points. We proceed with Hessian matrix again. In order to calculate it, we need to find zxx, which is 12x squared, zyy, which is 24y squared, and zxy is simply negative one. Okay. Now my choice will be at first zero, zero point. That A is a matrix starting with mixed derivatives. Here I have two zeros because these derivatives, these two, two zero, indefinite, that's what I can tell. These are subtle points. Now I need to be very careful with calculating the values at the rest of the points. So I have x, which is one over two to the ninth-eighths power and I have y value which is one 11-eighth. Our zxx derivatives becomes two squared times three. One-fourth only, looks reasonable as a fraction. Now zyy or simply that's one-fourth. Now, we're ready to substitute into the Hessian matrix. Let it be a point in the first quadrant being B points. So D squared z add B becomes, we put minus one, minus one for mixed derivatives. Here we have add three. What we can tell if pulls it to blend in principle minor. Now calculation or the determinant gives us minus one, eight also positive. So B is a strict local minimum because it's positive. So B local min. What will change if we substitute C point with the minuses, a point which belongs to the third quadrant? We see that these derivatives and then they hype squares, so nothing will change. So C is also local min. The question that arises whether these two points can be points of global minimum. What's about global maximum, does it exist? We'll answer these questions immediately.