 So we have to take up applications of maxima and minima concept today, application of maxima minima. In fact, I'd already started with it in the last class when I discussed with you the nature of the roots of the cubic polynomial equation. So I will say application of maxima minima continued. Okay. So in this segment, we're going to talk about the optimization problems. To be more precise, nonlinear optimization problems. Not like the ones you have in linear programming problem LPP. So in optimization problems, I'll tell you what is the crux of these kind of problems. Here they will be an objective function. Okay. And you have to maximize or minimize that objective function subject to certain nonlinear constraints. So objective function itself may be a nonlinear function of the variable, which is changing. And you also have a constraint, which also is a nonlinear function. So on the basis of this nonlinear constraint, we need to maximize or minimize in short, maximization and minimization. The word itself means optimization and objective function. So we have to maximize or minimize an objective function. So what is the process behind, you know, carrying out this optimization. So let us say this is your constraint function. Okay. So let's say there are two variables involved. In fact, there may be many variables involved. So let's say there is a constraint function like this. Okay. This is the constraint function constraint function. Okay. And there will be an objective function. Okay. So let me, let me just write a generic OX1 for objective function. Okay. So you need to optimize this, optimize this objective function. So what do we do? We first restructure this optimal restructure these objective function in terms of a single variable. Okay. So many times there are multiple variables involved. Let us say I talk about a simple example that let's say there is a rectangle which has got some kind of a length and a breadth. Okay. Length is X, breadth is Y. And I say the perimeter of this rectangle is a fixed value. Okay. That means the length of the, you can say the sides are fixed. But still you want to maximize the area of that particular rectangle figure. So how do you do that? So in this, the constraint is that the perimeter is fixed. You cannot make the perimeter infinitely big. That is the constraint. And the objective function is the area, which is the product of let's say length into breadth. So it's a two-variable function. You need to maximize it. So what do we do in this case? We write one of the variables in terms of the other from the constraint and we convert it to a single variable function. So what do we do? Our effort is to convert it to a single variable function, if possible. Okay. There are many cases where you can manage without also. So we convert it to a single variable function. Okay. Let's say I call that function as a, you know, something like F, you know, X or Y depending upon the case that you have. And then in order to have a maxima or a minima for this objective function, the derivative of that function with respect to that variable in which you have written it must be zero. Okay. From here, we get the roots, which are the roots corresponding to the maximum or minimum value of the function. Okay. Let's say the roots come out to be alpha from here. One of the roots is alpha. We basically check whether the double derivative at alpha is positive or negative. If it is positive, that means your objective function f of X is minimum value or has a minimum value at X equal to alpha. And if it is negative, that means double derivative is negative. It means your function has a maximum value at X equal to alpha. Okay. So this is the sequence of steps that we follow. Let's understand it in more details when we take more questions. I'm sure most of you have already done problems based on this and school. So let's get started with some problem solving and we will learn more from the problems themselves, rather than talking about things in general like this. So let me begin with my first question. I wanted to copy this down. I mean you're already aware of the process is just that we have to take up some questions to understand it. So let's start with the very first question. Okay, let's take this question first. This question was asked in ITJ 1991. Normally there is, okay, yeah, that's a very good question. Can there be only one constraint function? If you ask me in general, need or not be, there may be more than one constraint function. Okay. But ultimately, whatever constraint functions you have, you need to write down your objective function in terms of one of the variables. Okay. So yes, there may be multiple constraint functions. But many times we see only one constraint function, two is very rare. All right, so all of you please read this particular question. It says that a window of fixed perimeter, including the base of the arc is in the form of a rectangle surmounted by a semicircle. The semicircular portion is fitted with colored glass while the rectangular portion is fitted with clear glass. The clear glass transmits three times as much light per square meter as the colored glass does. What is the ratio of the side of the rectangle so that the window transmits the maximum light? And sorry, I forgot to wish, wishing you all a very, very festive season of Burma Lakshmi and Muharram as well. These are Indian festivals. I hope you celebrated it during the morning time. Okay. Good. Good to know that. Whereas ours festival is studies. Isn't it? That is the major thing. Okay. So should we discuss this problem? I mean, would you like to try it or should I discuss it with you so that you get an idea how to proceed? How do you want it to go? Okay. You want to try it? Okay. Go ahead. Sure. Could you draw the diagram? Okay. Sure. See, this question has a rectangular structure of the window. Okay. And it is surmounted by it is surmounted by a semicircular part. Let me raise this one. Now, this semicircular is fitted with colored glass. So here you're putting a colored glass. Whereas this part has a clear glass. Okay. Now, what are the restriction here? Restriction is that the perimeter of the perimeter of this particular window is fixed. Okay. The perimeter is fixed. And they're also talking about including the including the base of this particular arc. Okay. So that perimeter is fixed. And you want the light to be maximum from this. Okay. A few things to be taken care of here is that the clear glass transmits three times as much light as the colored glass does. Okay. So the constraint is you may choose your dimensions as per your wins. For example, I may choose this length as X, X, and let's choose this S to Y. So what is given to us is that perimeter, which is 2X. In fact, 2X plus 2X, 4X plus 4Y plus half of this, which is going to be Pi X. This P is a constant. This P is a constant. So this is our constraint. The constraint is this is a constant. And the amount of light which is being passed by this, that is your objective function. Okay. So let us say the amount of light that passes per unit area from the colored glass is lambda. Whatever unit is there for luminous intensity. So lambda per unit, you can say area is the light passing through the colored glass. Okay. So lambda Pi X square by two, that would be the total amount of light coming from the colored part. Okay. And the amount of light coming from the clear part is three lambda times whatever is the area, area will be 2X into 2Y, which is 4XY. So now here I'll stop. Here I will leave up to you to do the working. So you have to maximize L under the constraint P given to you. Could you explain how you got that equation of L? You're talking about L. Whatever. Okay. See, what is the area of this semicircle Pi X square by two? And let's say lambda is some, you can say it is the amount of light passing through that particular area per unit area. Okay. So area into lambda will give you the total luminous, you know, intensity, which is basically crossing that particular part and three lambda because the clear part of this particular window transmits thrice the amount of light. Okay. Normally when we have fancy windows at home, you'll see that people will put two types of glasses. In fact, multiple types of glasses. Okay. So the glass which has been fitted on the top in the semi-circular part that is transmitting three times lesser light as compared to the clear part. So that's why this is lambda and this is three lambda times the area of this particular rectangular part. Okay. Okay. Okay. So I will also know pitching because it might take longer time to solve this. So as you realize that this objective function is in two variables. X is also there. Y is also there. Right. But what we discussed in the previous slide is that we must try to convert our objective function into a single variable function so that we can differentiate that function with respect to that variable. So here I plan to convert everything in terms of X only. So this four Y over here that you see, I will, you know, obtain from the first equation by using P minus pi X minus four X. Okay. That is four Y. And of course there was an X also there. So X will be multiplied to this. If you want, I can multiply X right here itself. Is it fine? Okay. Now for L to be optimum, whether maximum or minimum, I have to, you know, ensure that DL by DX. Okay. Now many people ask me, sir, could I have written everything in terms of Y also? Yes. It is your call. Okay. But write it in terms of one of the variables. In fact, you can use. There's no such rule that you have to always write in terms of one variable. If need be, we can actually work with two variables also. Okay. Yeah. So I will take up a problem based on that type as well. Okay. So now if you differentiate L with respect to X and put it to zero on this side, you will end up getting a lambda pi X. This I'm just splitting it out. It'll give me three lambda P minus six pi lambda X minus 24 lambda X. Okay. Correct me if I'm wrong. And this must be zero. Okay. So from here, from here, you can do one thing. You can drop the lambdas first of all, there's no point of keeping a lambda because lambda will easily get cancelled off. So you can say three P. These two will be collected as minus five pi X and minus 24 X is equal to zero. Okay. So from here, you get your X as three P by five pi plus 24. Okay. So if X is this. Okay. What is going to be your Y. So let us use our constraint given to us. So our constraint says for Y is equal to P minus. P minus pi X. So instead of pi X, I will write as three P by pi plus five pi plus 24. And you have pi X minus four X. So minus four X will be nothing but minus 12 P by five pi plus 24. Okay. So this will be your four Y. Let us try to simplify this a bit. I can take a P common and I can take five pi plus 24 common. So it will be five pi plus 24 minus three pi minus 12. Okay. So that is equal to four Y. So four Y is nothing but P times. This is two pi plus 12 by five pi plus 24. Okay. Now what we have been asked, we have been asked for the ratio of the sites of the rectangle so that the window transmits the maximum light. So what I have to find out, I have to find out the ratio, the ratio of the sites, which is let's say two X in this case by two Y. Okay. In short, you have been asked for X by Y. So X is this. Right. Y is going to be. This divided by four. So you can write this divided by four. So you can write P into two pi plus 12. I'm so sorry for writing X here. It is five. Yeah. And five pi plus 24 will be on top, which will get cancelled with this. PP will also get cancelled with this. Correct. And there was a factor of four also, which I forgot. So this is going to be 12 by two pi plus 12. The answer is six by pi plus six. Okay. Now here, one thing I have missed out, which you should not be doing. I have not done the double derivative check. And I've not figured out whether at, at this value of X does the amount of light become maximum or minimum. That test is very, very important. Now I cannot say it is very important for your board levels because even that has become objective. Right. But yes, you must always perform the double derivative test. Double derivative test. In case you are writing the same topic in a subjective paper. So double derivative of L with respect to X calculated at that X, which is three P by a five, five plus 24. This must come out to be negative. Okay. I think it's very easy. You can easily figure it out. So if you do a double derivative here with respect to X, you will end up seeing a lambda pi. This will go off. This will be minus six pi lambda. And this will be minus 24 lambda. Right. And this is going to be minus 24 lambda minus five pi lambda, which is definitely a negative quantity. Assuming my lambda is positive because as the amount of area is increasing, the light passing through will also increase. So this lambda is a positive quantity for sure. Okay. So this test is important. However, this test is not required to be done when you are solving competitive level exams questions, because it will take away probably, you know, 40 seconds or 50 seconds of your time. Okay. So this question came in. IIT J exam. 1991. I think none of you were born with it. By doing that. Is it fine? Any question? Any concerns? How the process is basically, you know, followed. So again, I'll repeat, write down the constraint, write down the objective function, try to write down the objective function in a single variable, differentiate the objective function with respect to a single variable. See, many people say, if I don't convert it to a single variable, then what we can do? What is the other option with you? The other option is let's say you keep it in both X and Y. Then you differentiate it with respect to one of the variables. Let's say X. Then at one point you will see there is a dy by dx coming up. So you can relate dy by dx or you can write dy by dx from the constraint equation. Are you getting my point? So you can replace that dy by dx from the constraint equation. So you can differentiate here both sides with respect to X. Okay. And write down dy by dx in terms of X and replace it back there. That is another way which people follow. It is up to you. Did you understand what did I say? Is the second approach clear? What did I try to say here? See, approach 2 is where let's say you differentiate it L with respect to X. Okay. So what will you get? You will get lambda pi X and here you will get see 12th lambda and you will get X dy by dx plus Y, isn't it? Now you differentiate this side also with respect to X. So dp by dx will be 0 and this will be 4 and this will be 4 dy by dx. Okay. So this dy by dx, you can find it out and replace it over here. That's another way to do it. Okay. Depending upon what is the need of the hour, you can follow this approach also. Is it fine? So I'm not doing it. Maybe you can try it out on your own as well. Okay. Let's move on to the next question. We will take few questions to understand the nitty gritties of the situation. Okay. Let's take this question, a simpler one. A right circular cone of maximum volume, which can be inscribed in a sphere of radius R has altitude equal to which of the following options. Once you're done, give me a response on the pole. I will be running the pole as well, but not now, maybe in few minutes time. Once you're right, meanwhile, as you're solving it, I will make the diagram. So this is the sphere under which you have to inscribe a right circular cone. Okay. Right circular cone is a cone where the axis of the cone is perpendicular to the base. Now, why can't this right circular cone have infinite volume? If you want to maximize volume, it can go to infinity, right? The problem is that it has a constraint. The constraint is it has to be within a sphere of radius R. That is the constraint, right? It's like in our day to day life, we have constraints related to so many things. We have constraints with respect to time. We only have a 24 hours day, right? So God has not given us infinite long day. So we have to make use of, you know, that constraint and maximize our objective. Okay. So this particular right circular cone has a constraint that it cannot go beyond this particular sphere. And it cannot be of any orbit shape. It has to be a right circular cone. Okay. So, let us first wait for people to give a response. Okay. Shrita has given. So let me put the poll on because... Oh, sorry. Poll has not been created. Anyways, you can give me a response on the chat box. Okay, Gayatri. Okay. I got that. I could figure out that symbol was right above 3 actually. Aditya. Okay. So two different answers so far. Okay, Pratik. Above. Okay. Let's discuss it. Now, many times we would require some extra, you know, a perimeter to help us solve the question right now. I believe that if I could take this angle as theta, my life will be simpler, right? So I'm not claiming that you cannot solve the question without it. But taking this as theta will really help you out. Okay. Now, here we have... We can directly write down the objective function. So our objective function is basically to maximize the volume of this cone. So let us write down the volume of this cone in terms of the radius of this particular sphere and this angle. Now, I can clearly see that the radius of this particular cone is given by... This is r. This is theta. So this is r sine theta. So it is r sine theta whole square. And height is nothing but r plus r cos theta. Okay. So you can say r plus r cos theta will be your height. So you can see here that your objective function is automatically written in terms of one variable. And here the variable is theta. So theta is the only variable because if you change the dimension of this right circular cone, theta is going to vary. Nothing else is going to change. So let me write it in a... Yeah. In a sanitized manner. Now here, the derivative of the volume with respect to theta must be zero if you want your volume to be maximum or minimum. Okay. So this condition is the necessary condition for both maximum and minimum. So this particular condition must be satisfied. So when you differentiate it with respect to theta, you will end up getting... This is nothing but sine square into minus sine, which is minus sine cube. And this is going to be one plus cos theta into two sine theta cos theta. Okay. This should become a zero. Okay. So from here you can say that sine cube theta is equal to two sine theta cos theta one plus cos theta. Okay. Now remember sine theta cannot be zero because if theta is zero, this particular right circular cone will collapse. Okay. Neither can it be pi. Pi also will make it collapse. Okay. So one of the sine theta's get cancelled off. Okay. And you have something like this. You can write your sine square as one minus cos square and one minus cos square is one plus cos theta, one minus cos theta. And you can cancel off this as well because you know cos theta is not minus one because for that theta will become pi, which is not possible. So this gives us three cos theta is equal to one. So cos theta is one third. Okay. Now please do our double derivative tests, which I'm not doing here. The double derivative of the volume with respect to theta at cos inverse one third, this should come out to be negative. Okay. So please perform this test. Okay. Internally. And now they're asking us for the height. So height is nothing but r plus. Height of the cone is nothing but r plus r cos theta, which is nothing but r plus r by three, which is four r by three. Option number C is correct. Option number C is correct. I think most of you got it right. Apart from. Aditya, what happened? You're the only person to get this wrong. Yeah, please check your, please check your working. Okay. Next question. Okay. So we'll take some. So this question says an electric light is placed directly over the center of a circular plot. Of lawn 100 meters in diameter. Assuming that the intensity of the light varies. Directly as the sign of the angle at which it strikes an illuminated surface. And inversely as the square of its distance. From its surface. Okay. How should the light be hung in order that the intensity may be as great as possible at the circumference of the plot. Read this question. Understand this question. What it is trying to say. Oh, sorry. Distance is from the light source. I mean, you would have done it in physics as well. Right. So the intensity of a light. Varies inversely square. Of the distance from the light source. Correct. So let's say this is the spot on the circumference. Let's say there's a stadium. There's a running track. Now you want the athletes to practice during the night. Okay. And they're running on the track. So at whatever spot the athlete is on the track, the distance of that. Distance of that point from the light source. The intensity varies as inversely square of that distance. Getting my point. So this distance is what you are looking for. This is your distance. Okay. And it varies as the sign of this angle. Correct. So the farther away the man from the light source, obviously distance is also increasing and sign is reducing. So the intensity of light will keep on diminishing. It's a very easy question. I mean, you just have to write down the expression and do yes. Yes. That's what that is the purpose. No. You want it to be as great as possible on the circumference. So every point will be at the same distance. That is why you are hanging your hanging the line in the, in the center of this lawn. See the question is directly over the center of the circular plot. Okay. So wherever you take your S. Right. All the points will get the same intensity. So you can just worry about one point being maximum. All the points will be taken care of automatically. Yes. Should we discuss it now? See, very simple. This is 50. This distance is 50 from here to here. Okay. So your D is going to be 50. Seek theta and the amount of light. Let us say the intensity of light that is falling is going to be proportional to let me write proportionality constant. Propositional to sign of the angle divided by square of the distance, which is 50 C theta square. Okay. You can remove the proportionality constant and place some lambda that is not going to affect our answer. Okay. So now you want your intensity to be maximum. So if I has to be maximum slash minimum, it implies that di by d theta must be zero. Correct. So first of all, writing this properly, it will be a lambda by 50 square. And this is going to be sin theta cos square theta. So di by d theta lambda by 50 square. You can write as it is. And derivative of this is going to be a cos cube theta minus sin theta and minus two sin theta cos theta. So it's minus two. And this must be equal to zero. This must be equal to zero, which clearly implies that cos cube theta should be equal to two sin square theta cos theta. You know, cos theta. Cos theta cannot be zero because for that theta has to be 90 degree, which cannot happen. Okay. Because for that this, the track has to be of zero radius. So cos square theta is equal to two sin square theta. In short, you can get your tan theta as one by root two. Okay. So if one by root two is this, what is the height? What is this height at which you should have hung this light source? The light is nothing but at a distance of 50 tan theta. Isn't it? So your answer will be 50 by root two. That's 25 root two meters. So the person or the stadium management should hang this light at a distance of 25 root two meters from the, from that track, you know, in the, up in the air. Okay. And again, I have not done the double derivative test, but if at all there is a subjective paper, you must check the double derivative at tan inverse one by root two to check whether this is negative or not. Please do this checking. I am not doing it. In the interest of time. Okay. It's just a manual work. Is it fine? Any questions? Any questions? Any concerns? Please note down anything. If you want to ask anything, please do so. Else you'll move on to the next question. Clear? Can I move on to the next one? Okay. I think they are missed out a minus. Next question is a geometry based question. So there are three points given to you. A, B, C, A is P square minus P. B is Q square minus Q and C is R square minus R. They are the vertices of a triangle ABC. A parallel gram, A F D E is drawn with vertices. D, E, F on the line segments, B, C, C and A, B, respectively. So, A is P square minus P, B is Q square minus Q and C is R square minus R. They are the vertices of a triangle ABC. They are the line segments, B, C, C and A, B, respectively. Show that the maximum area of such a parallel gram is this. Okay. Ideally they should write a modulus here. I mean, it's a good practice to write a model. Yeah. So read the question once again. Please ensure you have understood it well. In fact, I'll draw a diagram also for you. Okay. And there's a parallelogram A F D E, which is drawn with D on BC. So let's say D is here somewhere. E on C A, let's say E is here somewhere. In fact, I have to make a parallelogram, right? So I must ensure D is parallel to AB. Okay. And F is somewhere over here because I have to ensure DF is parallel to AC. This parallelogram will be constructed. So this is a parallelogram. This is parallel to this and this is parallel to this. You want the area of the parallelogram A F D E to be maximum. Okay. Let us try to solve this as it'll take a lot of time. See, let us write down some lens over here. Let's say I call this lend as X. Okay. And I call this lent as let's say Y. Okay. And this angle is angle A. Okay. So in terms of XY and A, what do you think is the area of the parallelogram? Just type that in your chat box. In terms of XY and angle A, what is the area of the parallelogram? XY sine A. Correct. Everybody agrees with it? Correct. In this objective function, what is constant A? Because the vertices of the triangle is fixed. What is changing X and Y? Okay. So you can have different types of parallelogram, which I, which can be made, but you have to make a parallelogram with the maximum area. Okay. So there are two variables X and Y and there's one constant A. So in order to differentiate it first, I need to convert it to a single variable objective function. Correct. How will I do that? In class 10s, you would have studied Thales theorem, basic proportionality theorem. That is what we call it. Right. So if the two sites are parallel in a triangle, let's say you draw it, then we can say that the ratio of the sites will be same as, let me write it down. So can I say in this triangle CDE and triangle CBA are similar? Right. So can I say this, that DE, which is also X. Okay. DE is also X. Please note this down. So DE by, let's say AB, that is your length, let's say small c. We normally call the side opposite to angle C as small c in a triangle. Okay. This is equal to, this is equal to, can we say CE by CA? Okay. And CA is what? CA is small b. Let me write it down here. Okay. Side opposite to. So basically I'm not touching the coordinates as of now. Many people will say, where are you using the coordinates? As of now I'm not using the coordinates. I'm just basically using my side lengths of the triangle to get my job done as far as I can. But mind you, small c, small b, et cetera, they will all be constants. Why they will be constants is because the vertices are fixed. So side lengths will also be fixed. Isn't it? So why am I doing it? I'm doing this to get a relationship between X and Y. Are you getting my point? What is the rationale behind it? To get a relationship between X and Y. So DE is X. Right. And AB is C. CE. CE will be, if I'm not mistaken, Y minus B. By CA, CA will be B. In short, you end up getting, correct me if I'm wrong. X by C is one minus Y by B. So Y is equal to B times one minus X by C. Okay. So this I will substitute over here. This is just to make my objective function P as a single variable function. So I'll get X. Y will be B one minus X by C sine A. Okay. In short, it will become B sine A. And this will be X minus X square by C. Now, in order to make your P maximum or minimum, the derivative of P with respect to X must be zero, which means B sine A, one minus two X by C must be zero, which clearly means this must be zero, which means X must be C by two. Correct. Convinced. If X is C by two, you will check that Y becomes, if you put C by two here, Y will become B by two. In short, the area of the parallelogram, in short, the area of the parallelogram becomes maximum when X is C by two, Y is B by two. And of course, sine A will remain sine A. In short, in short, it's half of, half of B C sine A. Who will tell me what is half B C sine A? What does this represent actually? What is half B C sine A? Right. It's the area of the triangle A B C. Now here, the role of coordinates will come into picture. Okay. Now, how do you find the area of a triangle when you know its vertices? Everybody knows the formula. Area of a triangle, when the vertices are known is half X one, Y one one, X two, Y two one, X three, Y three one. Correct. And most of you have done determinants also. I didn't get that. Could you try again? Okay, Siri. Just shut up and get lost. I don't know why my computer wants to participate in the class. I don't know. Okay. So what is X one p square? If I'm, if I could just refer to the coordinates once again, give me one minute. Yeah, p square minus p one q square minus q one r square minus r one. Okay. If you have all done determinants in your school, can I use a property of determinant by which we can take one row and subtract it from the others? Have you done that property in school in determinants? Yeah, I'm sure you must have done. Rashmika. Thank you. Thank you for confirming that. So what I'll do is I will transform r two as r one minus r three and r two as r two minus r three. So I'm taking my third row and properties is deleted, but still these, these concepts you must have done. What among us hasn't done it? Oh, you're Rama. Aditya P. Oh, Aditya means DPS, right? Oh, HSR has done it. Okay. What should I do? Okay. Let me teach you something here before I solve it. What about gear international? Gaurav. Yes, you can use, but you'll get an ugly expression. Gear has done it. Gear SSRVM, et cetera. Believe it or not. Fast, fast, fast. Okay. Gaurav has done it. SSRVM. Anybody from SSRVM? Aditya is there. No, Aditya is not there. Easy. Let me teach you this. This is not a rocket science. See, there is something called row transformations. If you take any row and subtract it from the other rows, then this doesn't change the determinant. Okay. Again, there is a story behind it, which I will take when we start with determinist topic, which immediately after this we are going to do, but not today. In fact, today we are just going to start with matrices. Okay. So anyways, the answer here would be half. If I do this operation, it will be P square minus R square. This will become R minus P zero. Q square minus R square. R minus Q zero. R square minus R and one. Okay. And you can take your P minus R, etc. common out from first row. So if you take P minus R common, you'll get P plus R minus one zero. Take you minus R common from second row. You get Q plus R minus one zero. And now you can expand it. You can expand it with respect to column number three. So if you expand it with respect to column number three, this is what you end up getting. Yeah. Negative of this plus this. So it's Q plus R minus P plus R. So minus P minus R. So it'll be just getting Q minus P from it. By the way, in order to keep your answer positive, you must take a modulus of this entire thing. So it'll be half modulus P minus R, Q minus R and Q minus P. So this is the area of the triangle. Okay. What I've been asked to us is the area of the parallelogram, which we figured is going to be half of this, right? So this answer is going to be, I'm writing it in different color. It's going to be half of half, which is one fourth of modulus P minus R, Q minus R. P minus R, Q minus R and Q minus P. In fact, this is what we had been asked to prove. This is what we had been asked to prove. Yeah. Please ignore the order in which they have written because it has been modded. So P minus Q, Q minus R, R minus P. Okay. R minus P only you can switch it off. Okay. So this is how you solve this question. So here you can see that the coordinates were involved at the end, not in the, in the beginning. Is it fine? Any question with respect to the way I solved this question? So I wrote the area of the parallelogram in terms of X, Y and sign of the angle A, then I got a relationship between X and Y by using my basic proportional theorem. Once I got it, I differentiated it, differentiated the objective function, put it to zero. I got my X and Y, which I realized that we are such that it gave you the maximum area of the parallelogram as half the area of the triangle. Then I use the coordinate to find the area of the triangle and I halved it and I got the result. That we'll discuss when I do determinants, Aditya. In fact, if you have one, one, one in, you know, any one of the rows or a column, you can take any one of them to subtract with the other two. So there's nothing like, you know, you have to always do R1 minus R3. Take any one of them. Take R1 subtracted from R2 R3. That will not make a difference to your answer. Okay. Any questions? A regular base, a regular square-based pyramid is inscribed in a sphere of given radius R so that all the vertices of the pyramid belong to this sphere, means lie on this sphere. Okay. Find the greatest volume of the pyramid. I hope everybody knows the fact that volume of a pyramid is one third area of the base times height of the pyramid. And this is applicable even to a cone. Cone is also a pyramid, right? So half area of the base in a cone is pi R square. Height is H. So this is valid even for a cone. So in this case, your pyramid has a square base and it has been inscribed in a sphere of radius R. So they're asking us for the greatest volume of the pyramid. Let me make a diagram. Basically, I'm showing the periphery of the cone. Drawing may not be very good. So please bear with me. So basically, I've shown that these four square base, this is your square base. Let's say I call it as ABCD. They're hitting against the surface of the sphere. Okay. So the part where they're hitting the surface of the sphere, I have shown it with this dotted circle. Okay. And this is the other vertex. You may call it as point, let's say. Oh, awesome. Very good. Very good. Let's say this is our center of the sphere. So far, Aditya has given me a response. Let us see whether that is right or wrong. Okay. All of you, please pay attention. I'm going to slice this sphere along with the pyramid along the taking a plane which passes through E and the diagonal AC. In short, my plane is like this. So this is a plane. This is a plane which is cutting the sphere and it's passing through the vertex E and the diagonal off and the diagonal off the square base. I hope you can imagine that. So I took a plane and I sliced the sphere in such a way that my plane is passing through E and the diagonal of the sphere. Okay. Now, why did I do that is because I wanted a triangular structure like this to work on my, so let's say this is your distance. This is your radius of the, this thing, sphere. Okay. So now if you see this particular cross-sectional view, so if you see from this direction, so you realize that you are going to see something very similar to what you saw when we are dealing with the cone question. A little while ago we took a cone question, right? So the way you dealt with the cone question, the scenario will become more or less similar, but of course the objective function is going to vary a bit. I made this thing in gray color, so I'll continue with the gray color only. Let me choose a triangular structure only. Okay. Now please understand here that this length that you are seeing is the length of the diagonal of this particular right circular cone, diagonal of the base of the pyramid. Are you getting my point? Okay. So let's say this is your R. This is also your R. This is also your R. Let me drop a perpendicular. This will be let's say an angle theta. Okay. And this is your AC. These two vertices AC and E is on the top over here. I hope you're able to imagine how this dissection or how this, you know, cutting off the plane has happened. Understand the figure. Then only I will proceed further. There's no point proceeding further without understanding the fact. Clear. Any questions? Okay. So you realize that EM would be the height of the, the height of the pyramid pyramid would be EM. Okay. And that happens to be capital R. And this length is our cost theta. So capital R plus our cost theta that happens to be the length of the pyramid. Okay. Now what is MC? MC is half the diagonal. Right. So MC is R sine theta, which is half the diagonal, which is half the diagonal of the square base. Are you getting it? So this, when you drop down, it will drop here at this point M will be here. So this is your half the diagonal. I hope you are able to relate to this figure. Both these figures you are able to relate. One, of course, is a 3D figure. I have tried to make one. And this is a 2D figure. So MC is half the diagonal square, half the length of the diagonal of the square base. So diagonal itself is to R sine theta. Okay. Now, normally let's say if this length is small a. Okay. Then root to a is your length of the diagonal. So let a be the square side length. Okay. So this is a, a, a, a. So root to a is 2R sine theta. So a will be root 2R sine theta. Okay. Now, having known the side length, what is the area of the base? Area of the base will be square of this, which is 2R square sine square theta. So what will be the volume of the pyramid? Volume of the pyramid is one third area of the base into length. Right. Yes or no. So dear all, this is very similar to the expression, which we saw a little while back. So I will save my time here and I will directly write down the result. So if this has to be zero. It means cost theta is one third. Check your result. We had done it, you know, in the, in today's session itself, a little while back. Okay. So once cost theta is known to you volume of the, I'm so sorry. Volume of the pyramid will be 2 third R cube sine square theta will be one minus one by nine and this will be one plus one by three. So your answer will be two third. This is going to be eight ninth. This is going to be four by three R cube. How much is this 16 into four 64 by 81 R cube, 64 by 81 R cube. And please also check that the double derivative of this with respect to theta will be negative. You can please make a check out of it. I will not do it. And I think let me see what was your answer. Aditya, absolutely correct. Well done. Aditya, did you find out an easier way to do it? Because you give the answer very fast. Any approach that you would like to share with the class that you took? I took theta the other way. I mean, the approach means did you, did you take the cross section itself, et cetera? I mean, if you could unmute and talk. Yes, yes. We can hear you. Aditya. I took theta the other way. I thought it would be easier to work. From university I thought it would be easier. But the approach was the same. Okay. So you took this angle as theta here. This guy is theta. Yes, sir. Yes, sir. Thank you. Thanks, Aditya. Is this fine? Any questions? Any concerns? By the way, I have not done the derivative and solve for theta, but since this expression was already known to us, I utilized that and saved my time. Is it okay? Could you repeat the path after differentiation? See, after I differentiated it, which I did not show here is it done because I had already done this part when I was dealing with the cone question. I think you were there in the session, right? So just check that part. We had derived that cos theta will become one third. So I use that directly over here to get the maximum volume. I use that. So sine square is one minus cos square. So one minus one third square and one plus one by three. Simplification. I did. That's it. Is it clear? Any questions? Okay. Let's move on to next question. Yes, we'll take this question. This came in ITJ 2015. Okay. Read this question. Everybody. I'm using my artistic skills here. So basically it is a question where you have been given that there is a cylinder. You can say there's a flask kind of a thing, which has got a fixed inner volume. Okay. So this volume is fixed. Okay. This is fixed as a V meter cube. V millimeter cube. And it is two millimeter thick wall. So this, this thickness is off two millimeters. And it has got a disc. This is a disc at the bottom. This thickness is also two millimeters. Now read the question carefully. The question says, if the volume of the material used to make the container is minimum, when the inner radius of the container is 10 mm, find the value of V by 250 pi. Now up till now what used to happen, it used to happen that we used to find out the value of the variable for which the objective function is maximum. Right. Here, the question has been designed in a slightly reverse manner. They have said that it is maximum when the variable value is 10. And they're asking you for the fixed quantity values. So V is actually a fixed value. As you can see, they've already said it has got a fixed inner volume. So the question has been designed in a reverse psychological manner. So instead of asking you for the value of the inner radius, which in this case is a variable, they have said that when the inner radius is 10, then the volume required to make the material, not the volume inside. So whatever material is required to complete the flask, that becomes minimum. Okay. By the way, I'll just remove this. This is just confusing the entire figure. So there is a, yeah, there is a solid disk down over here. Okay. So in order to make the volume of the material, I'll write it as VM to just distinguish you to be minimum. This R must be 10. So what should be the volume? What should be this fixed volume? So see the psychology in which they have played on. Right. So a person who has only practiced the other way round type of problem may find it slightly weird to read. That also, I mean, till now I was always solving for the unknown. Here they have given me the unknown. And I always had a known and they're, they're, they're asking you the known value or the fixed value. I hope I have been able to explain you the scenario here. Yes. Any progress? Okay. Okay. Good. Okay. Okay. Let us write down the expression for the volume of the material required. So volume of the material required will be nothing but first of all, volume of the disk down below whose area, whose volume will be Pi. Now let's say inner radius is R, but the disk radius will be R plus two actually. So Pi R plus two square into height, height is two. Okay. So this is the volume of the disk, which is down at the bottom of the flask. Now what about this annular cylinder? Annular cylinder will be having an area of Pi R plus two, the whole square H minus the volume inside. Isn't it? Let's say H is the height. Let's say H is the height of this particular hollow part inside the cylinder. Right. So Pi R plus two square H will be the outer part. And from inside, you remove that volume V that will be the volume of the material required. Isn't it? Now here there are two variables. One is H and other is R. Okay. So what I'm going to do is by the way, this V is a fixed value. It's a constant for us. So what I'm going to do is I'm going to write down my H in terms of R, which is very easy to write because V itself is Pi R square H. So from here, you can say H is V divided by Pi R square. Okay. So I'm going to replace that over here. Before that we can take, I think Pi R plus two whole square two plus H. Okay. Minus V. And this H. I'm going to replace with V by Pi R square. Okay. This is the volume of the material required. Volume of the material required. Is it fine? Any questions? Any concerns so far? Okay. Let me just do a quick simplification here. I will take R plus two whole square inside. Plus volume. Pi Pi will get canceled. So you have R plus two whole square by R square. Sorry. This, this will, this is already taken care of. So there will be a Pi. So far so good. Okay. So let's now differentiate this with respect to R and put it to zero. So Pi, this will be four R plus two. And this will be V by Pi. I can treat this as if you have written R plus two by R, the whole square, or you can say one plus two by R the whole square. So derivative of that will be two one plus two by R into derivative of this part will be minus two by R and this volume will be differentiated to zero. Okay. So from here, we end up getting four Pi. R plus two is equal to two V. One plus two R into two by R square. And this equation is satisfied. This is satisfied by R equal to 10 as per our question. See. Read the last statement here. If the volume of the material uses minimum when the radius is 10. So when R is 10, this is minimum. That means this condition is satisfied. So you put your value of, you put your value of 10 in your expression. So this will become 12. This will become one plus one by five into two by 100. So let's simplify this. So four Pi into 12. This is going to be two V. Six by five into two by 100. So I think this will cancel off the 12 part and this will cancel off the two part. Okay. So volume becomes 1000 Pi. Okay. Check it out. So in the question, they're asking you the value of V by 250 Pi because they wanted to make an integer type question out of it. So this divided by 250 Pi will give me an answer of four. Is this fine? So the reason I took this question was because I wanted to show you a different perspective of how a question can be framed. Okay. So with this, we conclude this chapter. I think you have already been given a worksheet on Maxima Minima, which has got word problems there. Okay. You can try that out for your assignments and now I'm going to begin with a new concept.