 Let me give you an example of an approach to optimization which is a kind of classical approach that does not use any modern arguments. So this problem is, it has known by several names, one name for it is what is called Queen Dido's problem, another name for it is what is called the isoperimetric problem. So the problem is the following, you have been, you have given a rope, this is some rope you have been given of a certain length of a length say L, you have given a rope like this, you have to create a shape, you have to create a shape out of this, create a shape of maximum area. You have a plane on which there is lying this rope, you have to create a loop out of this of maximum area, that encloses the maximum area. So by shape, this is a mean a closed loop, that enclose, so the enclosed area has to be maximum. Now, so does anyone know the solution of this, it is a circle, why is it a circle? So you can of course use of, there is a mathematical way of doing this, but like what I want to show you is this is how primitive people did optimization without even knowing it. So here this is an, but the reason I want to show you this is not because to teach you how as a technique, but rather to point out something else, but anyway this is also a pretty argument. So let us just go through this. So first, so we want to create some shape, so suppose this is some shape that we create. So this is your start and end point. Now is there some way you can argue that this is certainly not optimal? So suppose I in addition to this rope, I also give you a rubber band. What you do is you take this rubber band, stretch it and fit it on top of the shape. Now how would that rubber band fit on top of all this? So I am just going to draw it with that, I do not know if I can, I think I can change color here, draw it with red. How would that thing fit? So it will be taught around these points this way, then it will kind of trace this shape here and then again it will become taught here and maybe taught here like this. Now look at these sort of points here, where the rubber band has become taught like this. Here the rubber band is basically a straight line. Now from this dot to this dot, the shortest path is the path taken by the rubber band, it is the straight line distance. As the black part which the rope has taken has to therefore be longer. So instead of keeping the doing this, if I simply allow, if I simply took the rope, let the rope follow the straight line path, then I would have used less of the rope and increase the area because now all this part would have become part of the area. So without, in the same rope budget I would have increased the area for sure. I have a rope budget of length L without violating that I would have increased the area. So by what does, what this means is the first, we have our first observation which is that we cannot have dents like this, this kind of a dent, this sort of a dent, this kind of a dent, all of these dents can be flattened out, we cannot have dents like this. So all these dents can be flattened out, this can be made formal. So a set, let us call this set S, a set S is said to be convex if for all x, y in S and all lambda in the unit interval, lambda x plus 1 minus lambda times y belongs to S. Now what is this as lambda goes from 0 to 1, what are we doing here? You have, you have taken some set S and here are 2 points x and y in it. As lambda goes from 0 to 1, what are we spanning here? We are spanning the entire segment joining x and y. So what this says is, so the set is said to be convex if for all x, y in S, the entire segment joining x and y belongs to S. So what our first observation which is in which we basically colloquially simply said that it cannot have dents, what it is effectively saying is that the shape whatever is the shape of maximum area must be convex. So our first observation is that the shape must be convex. Now here is the second observation. So you create some shape, now we know it has to be convex but we do not know what it looks like. So here I have drawn some convex shape. This is, suppose I do the following, I consider, so this whole perimeter is length L, so I start from here and I start from this end point, start from this end point and go, suppose this is my end point and I start from this end point and I go in this direction and I stop at the point where the perimeter, where the length has become L by 2. So half the rope is on one side, half the rope is on the other side. So this here, this length all of it is L by 2, the other half is also L by 2. So now any observation you can make about this? So if you look at the shape formed by one half, you can always replicate it as shape formed by the other half, correct? So now if the left half, whatever you want to call the left half or one half, if that has lesser area than the other half, then you can simply copy the shape of the other half onto this one and increase the area. So what this means is around this line, the shape should be symmetric, right? So if I take this line which joins the starting point to the L by 2 point, around this line, the shape must be symmetric, it must be symmetric if it is of largest area, so this is necessary. And now you have the third observation, so if it is symmetric then it suffices to just look at half the shape, so let us just look at half the shape and try to say something about it. So here you have the diameter or L by 2 line and here you have this other, here you have the, okay, now question, now is what should be the shape of this half, okay? So any ideas, okay, why surface? Symmetric does not automatically give that it is a surface, see it could have been an ellipse for example and both halves would have looked symmetric, no, but I have taken the L by 2 line, so I have taken a specific line, I am not arguing that it is symmetric around every line, so this should, you should think of sort of calculus free arguments, you know, let us not think of infinite symbols and so on, this is purely a geometric problem. So here is one approach, so for example, consider this some point here on the, on this line, okay, and I will and so let us just look at this triangle, starting from here till here, now the way it looks right now to me it looks like this angle is obtuse. Now just imagine I had a hinge here, I had a hinge here and I could move these two parts around this hinge, okay, these two parts that have been formed, I can move them around this hinge, when I move them what would happen? These areas, this one, this shaded area and this shaded area would remain the same, right, the sides of the triangle, what I am changed by moving it thinking of that as a hinge, all I am doing is changing this angle here, this particular angle, but I am not changing the areas that are highlighted, the shaded areas, they remain the same, right, so I can keep moving it, I am also not changing the lengths of these sides of the triangle, this side length, this side let us call this suppose A and this B, they also remain the same, correct. Now while I when I move this around, the shaded areas do not change, but the area in this triangle, triangle formed by these A, B and that dotted base, the area of that triangle will change, it may increase, it may decrease, okay, and you tell me what should I make the angle to get the maximum area, by 90 degrees, yeah, so the area of the triangle is half base into height, so let us take one of these as the base, let us take the base as A, so base is not changing when I change the angle, the height is changing and what is the maximum height that I can get, I can make the B, these other side perpendicular to it and that will give me height equal to B, right, so area of triangle, this triangle is half A into height and the height is maximum when it is equal to B, which happens when that angle is a right angle triangle and so this is why at most half A into B, correct. So the shaded areas do not change, triangle area is maximized when it is a right angle triangle, right, so total area is maximized when this is a right angle triangle, right. Now, here is the interesting thing, so now we, so this, what we have concluded is well the shape of maximum area must have a right angle, must form a right angle here, but then I did this by choosing an arbitrary point on the circumference and making imagining a hinge at that point, I can do this for every point, right, so what it means is all of these, the shape should be such that all of these, this, this, this, all of these should be right angles and for that to be a right and that we know happens only for a circle, right. So then if all of these are right angle, right angle triangles implies that this must be a semicircle and the total thing should be a circle, is it clear? So this is a completely calculus and algebra free argument of showing what we already know that of the solution of the isoparametric problem is that, is the circle, okay. So this argument is due to Steiner, I do not recall the year, it is due to Steiner, this is called Steiner's rule, Jacob Steiner, okay. So now, so there is something interesting that has happened in the result, one is that of course we have got this conclusion, but what there is something slightly more subtle that has also happened. So what was the logic that we followed if you think about it, what was the logic? We said a shape, consider a shape, okay. So suppose if this shape is not a circle, then it can, then there is a, we have, what we did is we said we presented a method of manipulating it, one was by removing dents, okay, second is by symmetrization and third is by introducing this hinge, we presented this method of manipulating it in such a way that its area can be increased, right. So what was the logic? And hence we concluded that the circle must have the highest largest area, logic is if the shape and the conclusion is, okay. Now, there is a problem with this logic, the problem is the following, I can use this logic to prove many kinds of things, such as for example, I can prove this theorem, the theorem is of all positive integers, the largest, the proof is, what is the proof? If one is, if I have an integer that is not 1, then I have a method of increasing it to get an even larger number, larger integer, method of increasing it could be say for example, squaring it and I can keep doing this and eventually conclude by this, therefore the 1 is the largest number. If it is not 1, then squaring it will increase the size, increase the number, so therefore 1 is the largest, right. So there is, you can see there is something, something has gone wrong somewhere, I have applied the same logic as I did for this circle problem. So this is obviously absurd and so the reason it is absurd is that we have implicitly assumed that there is actually a solution to the problem. So the problem, earlier problem was that you have a rope, you wanted to find a shape that maximizes the area, you have assumed that there is such a shape, likewise here we have assumed that there is such a thing as the largest integer, whereas there is no such thing as a largest integer. In both cases, the logic has a gap, which is that we have assumed or we have taken for granted the existence of an optimal solution. In the first case, there was no problem because even though we did not assume it, miraculously there actually exists an optimal solution. In the second case, the logic actually catches you red handed because you have implicitly assumed it but there is no solution, there actually is no solution. So what this means is we have to be careful. So this is why there is a need for a more rigorous and careful approach to optimization. So this is, so the purpose we are telling you the story is twofold. One is just to show you a very acute and elegant proof of this isoparametric problem. The second is also to is to bring to for the fact that one has to read the subject very carefully. What it means for a solution, what is the solution, when does it exist etc. are all questions one has to be one has to be one has to take into account okay.