Now it's time to form the Lagrangian. If we look closely at the constraints sets, we can tell that for each point in this set y-coordinate is always positive. So, y never turns zero, that means we can skip that inequality and work with the rest. Then, Lagrangian we will start writing with objective function, then the order of terms is an important. I can start with, for instance, I write Lambda one. Remember we subtract from the bigger the smaller. Three minus y plus, so this slant line. I proceed with that, the curve, parabolic curve and the final term that will be Lambda four multiplied by x. So, this is our Lagrangian. Now, we set first order conditions. We'll need a lot of space to fill in this place, dL over dX. Next, dL over dY, and complimentary slackness conditions. For instance, following the same sequence of the constraints, then I write Lambda one multiplied by three minus y should be zero, Lambda two be zero, Lambda three, and finally. Next. Here going backwards we put a star and star inequalities. How to deal with the system of equations and also inequalities? We have many inequalities, non-negativity conditions on multipliers, and also the basic inequalities that form the constraints set. We will follow some scheme. First of all, let us check whether there are any critical points over the Lagrangian function L within the constraints set, when we exclude the boundary points. So, we can call it, we are looking for them in a solutions of this problem. By the way, if you look at all complementary slackness conditions which we have here that are four them, if we choose any point which doesn't belong to the boundary of the constraints set, then all Lambdas should turn zero. So, that will be step 1 in our solution on the problem given Lambda one equals Lambda two equals Lambda three, Lambda four is zero. What do we get? Well, it's quite sufficient to look at the first equation. All this terms disappear, was left on the y-variable that should be zero. But we know from the inequalities and also it's always useful to look at the drawing that y should be positive always. So, never happens, impossible. That's the conclusion that's impossible, meaning no inner solutions. Now, we start dealing with the boundary points. Our boundary consists of four segments. Segment between a and b. So, step 2 checking border points or boundary border, boundary it's the same boundary points. I'll proceed here. So, that step two will be split into four subparts. So, firstly, from a to b, here we have x equals zero. So, all we know that Lambda four should take non-negative values, as for the rest still zero. Okay. How does it comply with the rest? Actually, we're looking at the first derivatives, dL over dX, dL over dY. So, from the second equation, we get x should be zero but we know that we already, know that x is zero. So, that is satisfied. Here, we have from the first equation, we get y, this is zero, this is also zero, and that's left, plus Lambda four should be zero, not possible because y is greater than one and Lambda four is greater than zero or equal to zero. Then, a conclusion, the same as earlier, impossible, never happens. Now, from b to c only one multiplier can take non-zero value, positive value which is Lambda one. As for the rest, they are zeros. Once again, if you look at the first equation, we have zero, zero, and zero, impossible. Next, from c to d, this piece of the straight line. Meaning that Lambda two can take even positive values, and the rest are zeros. What will happen this time? This time, if you look at the second equation, we have Lambda one is zero, Lambda two. That makes some sense. We need to explore it in more detail.