We're going to find the point P of Xy on the line. Y equals two. X plus three. That is closest to the origin. We will talk about some points here. Where is the objective function? Dfx? What is available? We are going to be controlling. We'll call it X. What function modules? Security. This problem. What are the coordinates of the point satisfying the statement of the problem. And how far is that point from the origin? We are going to sketch the line also and market the point P mm hmm. On that line. And also the origin. So the spread line is why it was to explore three. Yeah. Mhm. Okay. Okay. So we know that when we have X equal um zero. Why equal three? So we know at this point is on the line at this point here and when X equals one for example, mhm. Why goes five? So it's gonna be upward? And we can say that for example negative one to interview one. Negative one is negative, two plus three is one. So at this point here get to be on the line. So the line get to be something like this or less. So we can say that the point is somewhere around this on this line and we had the origin here. So we want to find the point there is closest to the urgent on this blue line that is closest to the origin. So it's going to be around. Remember that the minimum distance corresponds to the uh or total distance. That is the point that is closest the point on the line that is closest to the original form uh 90 degrees between the light, the blue line and the line joining the point that is closest to the origin and the origin. Yeah. Mhm. Okay, that is the geometric situations. So mhm. Uh we are going to write the distance between a point that is on this line and the origin. Remember that? The distance between the points X one, Y one and X two, Y two is giving us the square root of X one minus x two square plus white one minus way to square. Yeah. And we want to find the minimum of that function. But uh we can say that the minimum of dysfunction is a team where the expression that is inside this group is minimum. Mhm. So we can say that the problem corresponds to minimize the expression side the square root. First of all, we are going to write the specific function we have here specific function is the distance from a point X, Y Z are defined this equation and 00 the origin. And this distance is equal to x minus serious square. There is x squared plus y minus serious square. There is white square. Mhm. And okay, we can say then that the function D of x Y equals square root of x square plus y square is our objective function and Y equals two X plus three. That is a condition that the point get to satisfy. So we got to minimize this function. Yeah, subject to these conditions. That is a problem we have and we can say that this problem is equivalent to the problem of minimizing this. Yeah, sorry, here, I forget the square minimizing the is subject to these conditions. That is the function square root of X squared plus y squared is going to be minimum when x squared plus y squared is minimum. So this is the optimization problem. We have this one, minimize this function subject to these conditions. And we can reduce this problem to the problem of one variable by using the fact that why I got to be equal to two X plus three. So why it was two X plus three implies that the can be viewed as a function of one variable eggs which is available. We are controlling the X coordinate of the point on the line and D. Is going to be equal to X square plus to explore three which is expression for Y square. So this is X square plus four X square plus 12 X plus nine, which is equal to five X square plus 12 eggs plus nine. So the problem has been reduced to minimize D of X equal to five X square plus 12 fakes plus nine subject two eggs on the line. Which is implicit in the fact that we use these condition for the second coordinated the point. That is we have in fact no restriction tina minimize this function and in this case X can be positive or negative. There is no restriction on that because the point had to be on the line. The important thing is that the point is on the line which is implicit in this substitution we had made here right, mm hmm. This one. So now to solve this, we get to find the derivative of tea which is 10 X plus 12. And distributive zero. If and only if 10 X is equal to negative 12, which is equivalent to X equal to negative 12/10. That is X equal, yeah, negative 6/5 mm or negative 1.2 if we want. So we can see that X is negative value. And that is consistent with the fact that in this graph, which is not perfect. Give us an idea. We saw that the point where we must have the perpendicular distance to the origin is a point with a negative coordinate X coordinate, so why must be positive instead? And that we will see later. But for now we're going to just to justify that. This is in fact the minimum value of the And we can say easily here because we can recognize here Parabola opens upwards because the coefficient, the coefficient of X square is positive. So this corresponds certainly to a minimum value of T. But we can see another way is to find the second derivative which is 10, which is always positive independently of the value of X. So this is a concave of function like this. And there is a minimum value for this function which is attained at X equals negative 6/5. Yeah. So we can say, yeah, he is the value where the of X has uh minimum and this is a global minimum. That is an absolute minimum because of these properties. So here function is always concave up. Yeah, that means there is only one minimum in the whole domain on the real numbers. So uh now we find why using the equation of the line, that is two, X plus three. Okay, Why equals uh two X plus three, That is two times negative. 6/5 plus three. That is negative 12 +45 or three. That is negative 12, 15/5. That is three or five is a positive value. So on the graph, yeah. Then the point Yeah. P Yes, negative 6/5, 3/5. That is 1/5 times upon eight of 63 Which is um where the distance to the origin or we're going to see better is the closest point to the origin on the line. Mhm To pds the closest point to the origin which is on the line, y equals two x 43 And now we want to find how far this point is from the origin. Okay. In that we find simply by calculating the distance from this point of the origin. No mhm 96 of the five 3/5 to the origin. And we saw that this square root. Now we got to take this because the square root, because we are calculating the actual distance. So we get to use the formula to distance that is the square root of negative six of five minus zero square plus three of the five minus serious square. That is square root off six square over five square plus three square over five square. That is one or five square root of six square three square. Uh That is 1/5 squared of 36 plus nine. One of the five squares of 45 mm and that is equal to 1/5 square root of nine times five square root of 93. So it's 3/5 square to five pes three over 500 to 5 apart from the original. So this is the minimum distance possible to the origin from points lying on the line Y equals to explore three. So that's the point place around here. And uh as we can see this problem was originally originally an optimization problem with two variable foot, we can reduce it to an optimization problem problem of one variable using the condition from which we can solve for one of the variables and stay with only one variable. And besides that, we have made the simplification of taking the radical expression instead of the whole expression with the square root. Because we know that the function with the square root or the distance is going to be minimum if the expression inside the square root of minimum. So we have done this to step. So originally disease the optimization problem. This is the first simplification, making the observation that these two functions has the minimum value at the same point. And then we have reduced the problem to one variable. Using the condition where we can solve for one of those virus. And finally it was solving the optimization problem in one variable, we find that the point that is closest to the origin and which lies from the line way to to explore. Three is the point uh negative 6/5 and 3/5. That is x is negative 6/5 and y is three over five. And the actual distance from that point to the Aborigines, three square to 5/5