So, by subtracting respectively the first three equations from one another, we get a system where the equation's affected. So, all these three equations are made of products. That allows us to split the solution of this big system, so into subsystems. Let's consider them. First of all, we need to explore three different cases. The first case being if y equals x, this equation is satisfied, but at the same time we know that when y equals x, both y and x should equal z because then if y, x, and z is the same number, that contradicts the constraint set. That means that we can consider the case when x equals y and not equal z. Then the third and the second equations are satisfied when both x and y equal the same number which is negative two lambda two. So then we have x equals minus two lambda two, y equals minus two lambda two. What about z? We can find z for the second equation. So, z equals four lambda two. So, what's missing? I'll include the final, the fifth. All I need to do, I need to substitute these values into the final. That becomes, when we substitute we get squares of lambda two squared. We have four, another four and 16 and that equals one. That gives us two values for lambda two. Lambda two equals either plus or minus one over the square root of 24 or this is plus, minus. That gives us two solutions by the way depending on the choice of the sign, either plus or minus. So that's how we get if we choose the plus, then we have a point, which is. Now, changing plus into minus provides us a symmetric point. All we have to do, we change all the signs here. Let's hope we've got two points. But we can get another, how many? Another four. When we equate successively we equate this time y and z, and x is different. So, if we equate y and z then choosing pluses here, and here we have with a negative sign, a double value. Another point, and when we change all the signs. So, what's missing? This time let us equate x and z, and the sixth point changing the signs from the previous case. So, that's how we got all six critical points or the Lagrangian function. We are looking for the maximum value or the function whose u equals x, y, z. Now, it's clear that when we multiply two negative values of x and y and then by a positive z, we get a positive value. So let it be point A. So the value of the function at point A equals two and here we have, we can get the same results. Let me put a tick here, a tick and also this one. So, out of six points starting with the function takes the same value. So, the assumption is that these three points that would be B and C are the points of maximum. How can we check that? We can employ various truss theorem because if we look at the constraints set we can visualize this set of points in 3D. So, this is an equation of a unit sphere and this is an equation of a plane which passes through the origin. So, this plane crosses this sphere and we get a circle. This circle is considered as a bounded set and it's also is a closed set, so combined. The boundedness and the closeness of the set gives us the compact. So, the function which is continuous on the constraints set takes the greatest and least values and we have found the greatest value of this function. So, the function takes this value at all three points, A, B, C.