 Let me begin with the first application which helps you to know the nature of the roots of a cubic polynomial equation. So I will be talking about one of the applications of maxima and minima which helps you to know the nature of roots of a cubic polynomial equation. So nature means how do you know what kind of roots does a polynomial equation has. That means how many real roots, how many imaginary, it also tells you how many repeated roots could be there. So all those concepts can also be figured out by the use of maxima and minima. Indirectly you can use calculus to know the nature of the roots of a cubic polynomial equation. So let us say I have a cubic polynomial equation like this. A x cube plus a x square plus b x plus c equal to 0. Now many people ask me sir why not a x cube b x square c x plus d, why did you choose your a as a 1. See it doesn't matter. Your roots position will not change even if you divide throughout with a. You anyways know in a cubic polynomial your leading coefficient will never be 0. So I am dividing throughout with that a. For the simple reason that I have to deal with lesser number of unknowns. Instead of dealing with a, b, c, d I choose to deal with a, b and c only. Just remember your nature of the roots is not going to change if you are going to divide the entire polynomial by a given non-zero constant. So let's say I want to know the nature of the root of this given equation. So let me call the polynomial on the left hand side. So in this polynomial equation this is your polynomial. So I am calling that polynomial as f of x. Now what do I do? I find its derivative. I find its derivative. And when I do its derivative I get a quadratic and there is lot of things which is dependent on this quadratic. In fact, lot of things will be dependent on its discriminant. Okay. Just a second. Yeah. Sorry. The discriminant of this quadratic equation plays a very vital role. So what I will do? I will take three cases of its discriminant. One, when your discriminant is negative. So I will start with the fact that the discriminant is negative. Now for a quadratic equation whose leading coefficient is positive, whose leading coefficient is positive, okay? So three is already positive and your discriminant is negative. What does it imply? It implies that f dash x will always be greater than zero. If your f dash x is greater than zero that means your f of x is an increasing function. So an increasing function and especially when it is a cubic function and it is increasing, it can only cut the function at one point, okay? So the function will only be able to cut at one point. So basically it will be like this, okay? It will never fall anywhere. It is always positive. So anywhere on this curve if you take the slope it is always be greater than zero. Right? So what does this imply about the, or what does this tell you about the nature of the root of this cubic polynomial equation? It tells you that it will have only one real root. One proper root. Yes. It will have only one real root. So it implies that f of x equal to zero will have only one real root. That means other two roots are imaginary because in order to create another root it must come down and cut the x axis again which cannot happen if it is always increasing. It cannot come down. Once gone up, it is gone up. It will not come down. Okay? So if it is not coming down, no more real roots will be generated. Okay? Yeah. Sure Shashank. I'll just call up. Shashank wants to copy something. Okay? So this is the starting point Shashank. You can copy from here. Now let us take the case, let us take the case where your discriminant is greater than zero. If a discriminant is greater than zero, what does it mean? It means f dash x equal to zero has two real and distinct roots. See, I'm talking about f dash x equal to zero. Please don't get me wrong. I'm not talking about f of x equal to zero. So if the quadratic discriminant is positive, we can conclude that the quadratic equation that is this equation equated to zero will have two real and distinct roots. If it is two real and distinct roots, can I say this quadratic over here can be factorized like this. So x1 and x2 are the two roots. I'm assuming here for the sake of simplicity that x1 is lesser than x2. Okay? Let x1 is lesser than x2. So x1 and x2 are the two roots, two real and distinct roots of f dash x equal to zero, not f of x equal to zero, f dash x equal to zero, okay? Now what does it mean? It means five situations can happen here, okay? So what are the five situations? Let us look into it. First situation is your function f of x can show this nature. You can go up, stop at x1, okay? Come down, stop at x2, again go up, okay? So as you can see the cubic equation or the cubic polynomial f of x can show this characteristic. So x1 and x2 are the points where the derivative becomes zero. Remember x1 and x2 are the roots of this equation. Let me write it as two real and distinct roots. Let's say x1 and x2, okay? So at x1 and x2 it will have a maxima and a minima, okay? That's why the derivative becomes zero, okay? So this could be one situation. Another situation could be, another situation could be like this. Let's say it goes up, comes down but doesn't cut and goes back again up. That means this is your x1, this is your x2, okay? This is your second situation, okay? Remember this is the graph of f of x which I am plotting, not f dash x anymore. Cubic polynomial graph. Third situation is this. Your function is coming up like this, not cutting and again going back like this. So this is your x1, this is your x2, okay? Let me call this as a third situation. Fourth situation could be like this. It could come like this. Let's not pass through origin unnecessarily, okay? Goes up, comes down, touches here and goes back again. So this is your x1, this is your x2 point, okay? This is your fourth situation. First fifth situation will be like this. It comes, touches and goes up like this. Can you think of any sixth situation? Have I missed out any situation? Let me know. What I can think of is these five situations can happen with the f of x polynomial. So this f of x polynomial will show any one of these five characteristics. So what's the fifth situation again? This is the fifth situation. Anything that I have missed out, do let me know. So please have a good look at it and tell me if I have missed out any situation. Okay. Now in the first situation, what is happening actually, you would realize that f of x1 will have an opposite sign as f of x2. That means f of x1 into f of x2 would be negative. Check it out. Yes or no? f of x1 would be a positive value over here. f of x2 would have a negative value over here. Correct? That means f of x1 into f of x2 would be negative. Okay. Now many people ask me, sir, why do you say this is opposite in sign? Why don't we just say f of x1 is positive and f of x2 is negative? Actually I can say that but the problem is here if you divide, say it all depends upon what was your leading coefficient. Correct? Are you getting my point? So when you divide by that leading coefficient and you get this, on this you can comment but on the initial equation you cannot comment. Okay. So what I can say is this is fine or you can say f of x1 is positive and f of x2 is negative. That is fine when you know your leading coefficient of the polynomial equation was 1. But this method which I'm writing, this is equally applicable even if your graph was going the other way around. For example, if your graph was going like this, so in this case what will happen? At x1 it is negative and x2 it is positive. So still their product is negative. So I want to account for both these type of polynomial equations or both these type of polynomial graphs. That is why I'm not putting any kind of extra restriction that f of x1 should always be positive and f of x2 is always negative. No. I don't want to put that restriction. Okay. Now when you realize that the value of the function at the roots of the quadratic that you got over here, x1 and x2 is giving you opposite sign then this is an indication that the polynomial equation will have 3 real and distinct roots. Okay. As you can see one root is here, another root is here, another root is here. Okay. So please make a note of it. Situation number one is a case where you realize that the polynomial equation will give you three real distinct roots. Situation number two is a one where you realize that f of x1 and f of x2 have the same sign. That means f of x1 into f of x2 is positive. And this is a case when you have again one real root. Two other roots are imaginary. So as you can see it has only one real root over here. So now you can see one real root. There could be two situations. Either you have f dash x always positive or you have two distinct roots of f dash x, but such roots for which f of x1 and f of x2 product is positive. And let's take a little bit further here. Here also you would realize that f of x1 into f of x2 will be positive. So this also gives you only one real root criteria. So there are three real roots criteria that you can see over here. That is situation number two, three and the first situation over here. Are you getting my point? In situation number four you realize, yeah somebody is saying something. In situation number four you will realize f of x1 into f of x2 is actually zero because here f of x2 is zero. In such case, in such case, what do we conclude? We conclude that it has got three real roots. Three real roots out of which one of them is repeated. Or you can say two repeated roots out of that. In this case, which is that repeated root? X2. So one root is here, let's say call it as some alpha or something. And the other root is x2, x2. Are you getting my point? So one root is different and two other roots are repeated and that is actually x2. Fifth situation also, you will have f of x1. Let me write it in white only because I have been writing in white so far. Fifth also f of x1 into f of x2 would be zero, which means there are three real roots out of which two are repeated. And that repeated root here is actually your x1. Here it was x2. Is that clear? Any questions regarding this? Out of this, this situation is very much commonly asked. The first one, the first situation. This situation is very commonly asked. I'll put a star next to it. Please copy this and let me know if you have any concerns. Yes, Trippan. Sir, I think this is a little off topic, but then like when we have some minima or maxima occurring at the x-axis, then how many, then we say that it's only two repeated roots there or we can say any other repeated root. Now, if it is touching there, it will have always repeated roots there. How many? Like here it's cubic so only two are possible, but if it's a higher... Yeah, there you cannot comment. There you cannot comment actually. Okay. If it is touching, it can have let's say x to the power four equal to zero. It has got four repeated roots there. So it is touching the x-axis at that point. So by the fact that it is touching, it need not be only two repeated roots. It can be more than that. So maybe we can save the degree of the polynomial as well and the other roots that exist. Like for example, if we have all the other roots touching for n minus two roots touching, then we must have two roots and stuff like I think that depends on the graph itself. That depends on imaginary roots and all. That depends upon the degree. So then in that case you won't have imaginary roots possible? Why not? We can have. We can have, no. For example, a six degree polynomial can have four repeated roots and two roots are imaginary. Yes, so we can't tell those, we don't know those two imaginary roots. We can't say unless I see if I tell you that there's a graph which looks like this. Now I don't tell you the degree of this polynomial. Let's say I don't tell you the degree of this polynomial. I mean you can make out from the graph that it is a fourth degree or something like that. But let us say if I give you such a graph by which you are not able to decide how many degrees is there. So if I just ask you how many places it is touching over here. So it could be two, it could be four also. But here you know it is going to be two only because one root is here and other root is here which are imaginary. So it has got two repeated roots and two imaginary roots. But if I just give you a graph like this, no. Okay, then probably you will not be able to tell what is the degree, without the degree you will not be able to tell how many repeated roots it has. Like x to the power four, x to the power six. Yes sir, yes. So those two whenever there's some, some f dash x equal to zero occurring somewhere else then that's an imaginary root. Where, where, which one? Sir, like in the graph you drew you marked those two as imaginary roots. Yes, yes, yes. So basically two roots are imaginary because it is not able to touch. Okay. Okay, so one root is real here and two roots are imaginary. Right, similarly here also. One root is real over here and two roots are imaginary. Okay. So moving on to the third scenario where your discriminant is equal to zero. When your discriminant is equal to zero, what does it mean? It means f dash x equal to zero has repeated roots. You can say has real and equal roots, which is actually repeated roots. So let us say your f dash x is factorizable as three x minus one x one the whole square. See, let's go back. Your f dash x was a quadratic. Check this out. Your f dash x was actually a quadratic. Isn't it? And if you're saying this quadratic has real and equal roots and let's say that root is x one. Can I say this quadratic I can factorize like this? Okay. So let's say that repeated root is let's let's say that repeated root is x one. Okay. So can I factorize it like this? So what does this mean? It means your f of x can be written as x minus x one whole cube plus a constant. I just integrated it. Correct. Now everything depends upon this guy. Everything depends upon this guy. If if k is positive. Okay. If k is positive or k is negative, so let me write it like this. If k is equal to zero, then we say that f of x equal to zero has three repeated roots, three real and repeated roots. And what is that real and repeated root here x one itself. So if you see this is zero, then it is basically becoming x minus x one cube whole equal to zero. Right. That means it has got three repeated roots x one x one x one. A classic example of this would be your x cube equal to zero. That means it is a perfect cube. If it is a perfect cube, then it will have three real repeated roots. Am I right? Yeah, sir. Correct. And if k is not equal to zero, then again your f of x has has one real root. Are you getting my point? Why has one real root? See, if I write x minus x one cube plus k equal to zero. So what will you write? This has x minus x one cube is equal to negative k. Correct. If you take the cube root of this, remember only one will be real out of it and two will be imaginary or non real. So as to say imaginary. Correct. Why am I writing half in yellow and half in? Yeah. For example, if let's say k was one, so negative one cube root, negative one cube root, you know it's negative one minus omega minus omega squared. So two of them will be imaginary. So out of this one real root, you can actually get your one real root for the equation. The other two will be not real. So they will be imaginary in nature. Even if your k was negative quantity, let's say even if you had a minus one over here and you brought it to the other side and made a one, cube root of one will also give you two imaginary quantities. Are you getting my point? So this just tries to say that the curve x cube will just shift x one right and k up or down depending upon its sign. So if k is positive, then it would the graph would be like this. So x cube graph shifted x one to the right and k up. So in such cases, it will generate only one real root as you can see over here. By the way, I should not make it pass through origin. It gives the wrong notion that it is passing through origin. So it is actually like this. So only one real root will be created over here. So this is the root. Okay. Even if a case negative, it will be shifted. You know, anyways, I mean it could shift right down right up. Okay. Left down left up wherever it shifts. It is like shifting off x cube graph. Remember your bridge course. If you are doing something to this function like this x minus x one cube plus k. What do you what do you infer from this? It means that x cube graph has been shifted. Yeah. Shifted by x one units right and k units up. Isn't it? Okay. That depends upon sign of x one and k also. So it could be like this also. So I'm not denying it. So this point could be your x one comma k point. So anyways, it will generate one root here. Is this idea clear? So please make a note of these three situations over here. When discriminant is less than zero, discriminant is greater than zero and when discriminant is equal to zero. Let's take questions on this and then we'll take a break. Can I go on to the next page? Yes, sir. Okay. So let's take a question. Find the value of a if this cubic equation cubic polynomial equation has three distinct real roots. In fact, they should say find the values of it. So find the values of it. We'll get a range of values. So which situation will you use over here? When is the only occasion when it has got three real distinct roots? f of x one into f of x two. When discriminant is greater than zero and f of x one, f of x two is negative in sign. Yes, sir. Done. Very good. Very good. Very good. See, if you differentiate this, let's say I call this as your function f of x. Okay. So if you differentiate f dash x, you'll end up getting three x square minus three. Okay. It clearly has two roots. Minus one and one. Correct. Okay. So let's say x one and x two are the two roots. So roots are plus minus one. So, okay, let's say this is x one and this is x two. Now, in order to have three distinct roots, f of minus one into f of one should be negative. f of minus one will give you minus one plus three plus a and f of one will give you one minus three plus a and this should be negative. That means a minus two into a plus two should be negative. That means a should lie between minus two to two. So this is going to be a ring interval. This is the answer. Getting my point here. Any questions? Let's take another one and after this we'll go for a break. Take the question carefully. It's a very good question. So there is a cubic polynomial equation in T. Okay. And later on this T is equated to x plus one by x. Okay. And they're saying it'll give you six real and distinct values of x. Okay. We'll see what is the right answer for that. Why you don't think so? Any reason for that? Yes. You have anything to tell me? Like by doing it, I think we get to non-cooperating intervals. Non-cooperating moment? Non-cooperating intervals. Like they don't, you know, exactly overlap. What are we mixing history and physics? Non-cooperation moment is happening, sir. Okay. Let's discuss this. Else it'll take a lot of time for us to solve the question. See, try to understand the question first. The question is saying that whatever T comes out from this equation, if you put it against the second equation. So let's say I get three values of T. Okay. So T takes three values, T1, T2 and T3. Okay. For each of these three values, if I solve this equation, let's say I solve this equation, I solve this equation, I solve this equation. Okay. Okay. I should get six real distinct answers for X. That means this should give me two answers. Okay. This should also give me two answers. This should also give me two answers because these are quadratic in X. And not only that, these answers should all be distinct. Different also. Different also. Correct. So for this to happen, first of all, T1, T2, T3 must itself be distinct. Correct. Correct. So first situation that I can say is that your T1, T2, T3 must be distinct. Must be real and distinct. And secondly, you would realize that if you want such an equation to have some roots, then your T1, T2, T3, all of them should be either be greater than two or less than minus two. Right. Secondly, your T1, T2, T3 must either be greater than two or less than minus two. Now, why is that so? Okay. If it is equal to two, you will have only one answer for it. That is one comma one. That is one and one. One is satisfied. Right. If it is less than two, you will not have any roots at all. Why? If you take a quantity x plus one by x and equate it to T, let's say. Okay. So if you just make a quadratic out of it, it will be having x square minus Tx plus one equal to zero. Right. So for you to have any roots, T square minus four must be greater than equal to zero. Correct. That means T must be greater than equal to two or T must be less than minus two. Correct. So if it is between minus two and two, there will not be any real roots at all. So forget about having six distinct roots. There will be no real roots at all coming from each of the equations. Now I've excluded two and minus two also because at two and minus two, they will have one one root each. Correct. So that means that will give me only three real distinct roots. In fact, that'll give me three real equal roots. That's why two and minus two also have excluded. Correct. Is that fine? Okay. Now taking into consideration all these things, let us say I go for the graph of this polynomial equation. So my graph should be such that my first root occurs before two. Right. So it is basically going up like this. Right. Yes or no. Yes. Second root should come after two and third root also after two. Something like this. Okay. Is this the only possibility or I could have more possibilities? Got all three be before minus two or after. Can all three be before minus two? Okay. So let's have a situation like this. What about the double derivative? Sorry, what about the derivative of this function? That will also tell you a lot of things. Let's differentiate this with respect to t. What do we get? Shankin, 60 square minus 18 t minus 18. Okay. That means your function maxima or minima should occur at zero and three. Am I right? So zero is here. Three is somewhere over here. Now, if you are claiming that all of them occurs before minus two or all of them occurs after minus two, then it will create a problem because it will not satisfy this particular condition. Yes. Can I say this is the only possibility? Sir, the root on the left can be after minus two also. It can be between, yeah, alpha one can be between minus two and two. But I think we need... Alpha one cannot be between minus two and two. These are your t values. These are your t values. Alpha one, alpha two, alpha three are your t values. Correct. So t values can be only be in this interval, no? It could be... But if you have to arrive at that of... Oh, okay. Okay. Okay. Yeah. So your root should be before minus two or after two and your function graph is such a way that it achieves a maxima and a minima at zero and three respectively. So this is the only possibility you can have. No other possibility. Correct. So now, with all this in my mind, let us try to get all the equations possible. So, can I say the first thing that I should adhere to is f of zero and f of three must be negative. Then only I'll have real and distinct roots. Okay. And at the same time, can I say f of minus two should be positive and f of two should also be positive? So second condition is f of minus two should be positive and f of two should also be positive. So this is my third condition. Okay. So my answer would be the intersection of all these three restrictions. So let us solve them one by one. So from the first one, what do I get? f zero into f of three should be negative. What is f zero? Thirty minus a. Okay. F three would be what? F three. Yeah. Yeah. F three would be two into 2754 minus 36 plus 30 minus a. Correct. No. Oh, wait a minute. So yes, three minus a correct. So this should be positive negative. In other words, you're saying a minus 30 a minus three should be negative. That means a should lie between three and 30. Okay. That's the first condition. Second says f of minus two. What is f of minus two? Less greater than zero. That I know. I want what is f of minus two. Excellent. Okay. I feel he is saying greater than two. F of minus two. So two into minus eight minus 36 plus 30 minus a. How much is it? Tell you are a J as parent taking so much time for doing simple calculations like this. So you're less than minus 22. That I don't want. I want the expression. I just want the expression. Are you saying minus 22 minus a you're getting? Yes. So a is less than 22. Okay. Third is f of two is greater than zero. What is f of two? A minus 10. Huh? 10 minus a. 10 minus a. 10 minus a greater than zero means a is greater than, sorry, a is less than 10. Okay. So what is the overlap of these three? Let's check it out. So the least values minus 22. Then we have three. Then we have 10 and then we have 30. First interval says between three to 30. Second interval says you have to be less than equal to 22. God. There itself it is gone. No overlap happening. Okay. Others is greater than less than 10. So there is no overlap. That means your a is. So when could you are correct? No solution. Was that a guess actually? No, sir. To be honest, I'd seen this question before. I didn't understand it, but I'd seen it in my school doubt group. Good. You're honest. So now we'll take a break. Right now the time on my watch is 649 p.m. I'm giving you two seconds extra here. We'll meet at 704 p.m. There we'll talk about optimization problems. Right. So I hope everybody's back. Yes, sir. So in this segment, we are going to talk about some optimization problems which are solved by the use of maximum and minimum. Okay. Now in optimization problems. Optimization means finding the maximum or minimum of some objective function. Okay. So what will happen is that they will give you a objective function. Okay. So this is what we call as an objective function. Objective function is a function which you want to maximize or minimize. Okay. And they will give you some constraints. Okay. And under this constraint, we need to maximize this objective function. Okay. So how do we do that? For that we have some simple steps that we follow. Number one, we first convert this function to a single variable function. We convert this to a single variable function by using the constraint function. Okay. After we have done that, we differentiate. We differentiate with respect to that single variable and put it zero. The roots of this equation will give you the extrema. Okay. And then the third step, which is very important and many students do not do this step in the school exams and they lose one mark or two. Is you have to verify the nature of the extrema points, whether it is actually giving you a maxima or it is giving you the minima. Let me explain this with an example. Let us say, let us say I asked you this question that if X and Y are positive real quantities that satisfy X plus Y is equal to 60. Okay. Then find the maximum value of X cubed Y. So now you can see here that your constraint function is this. So this is your constraint function. Okay. Of course, X and Y should be positive. So these are your constraints. And this is your objective function. This is what we call as the objective function. Objective function means it is your objective here to maximize or minimize this. That depends upon what is the question asked. So we have to maximize this objective function. Objective function. So what do we do? Step number one, as I've already mentioned, you convert this to a single variable. So what do we do? Step number one, as I've already mentioned, you convert this to a single variable function using the constraint. So what do I do is instead of this, I will write Y, I will write it as 60 minus X. Okay. So let's say I call this function as your function P. Okay. So now it is a single variable function. So this is now a single variable function. Single variable function. So step number two is now we differentiate the single variable function and put it to zero. So this is your step number one. Step number two, I'll add it over here. You differentiate the function with respect to that single variable and put it equal to zero. So derivative of X cubed 60 minus X, you have to find and put it to zero. If I'm not mistaken, derivative here would be 180 X squared minus 4X cubed, correct? This will be equal to zero. That means 4X squared 45 minus X will put it as zero. So there are two possibilities here. In fact, if you use your wavy curve, 45, you can write zero first here and 45. So this would be negative, positive and positive. Okay. So here we can know the nature of these critical points. So I get zero and 45 as my critical points. So 45 is the point where from positive slope it is becoming negative. This is the point of maxima. You can say local maxima. And zero is actually a point of inflection, which is a point of neither maximum or minimum. So X equal to 45 is a point where the function will achieve its maximum value. So if X is 45, Y will be 15. Okay. And then you can go ahead and find the value of this function, which is X cubed Y, which is 45 cubed into 15. That I'm not going to do. Okay. So this test is very important because when you get a point, there would be, there may be points of maxima and minima both, or there may be points of inflection also. So you need to tell the examiner that, okay, at this point a maxima will occur or at certain point minima will occur. So that test has to be done in the school exams. Sir, other than this wavy curve, the other test is double derivative. Double derivative. Yes. Yes. Okay. But if double derivative gives you zero for a point, that means you have to go for higher derivative tests. All those, or even and all that. Absolutely. Yes. Okay. So could you go to the rules for a minute? Yeah, sure. So C X comma Y is your constraint function. Done. Tirupan? Yes, sir. Thank you. Okay. So let us begin with a question. This, let's say there is a person, Venkat B. Okay. This person is trying to take a ladder from a point A to point B. Okay. And this ladder that he is carrying, let's say he is carrying a ladder in his hand like this. Okay. This ladder cannot be made vertical. It can only be kept horizontal. It is a heavy ladder. So you cannot make it vertical or it is, you know, such a ladder which balancing vertically is not possible. So you can only carry it horizontally like this. Okay. And this is a lane. This is the right angle lane. Okay. Now, of course, many of you would have also experienced this situation while shifting your houses or, you know, trying to buy a new furniture into your house, especially when you're living in an apartment with staircase. It becomes very difficult to maneuver the furniture about the corners. So let's say Venkat is trying to move a ladder. Let me make it a different color across this right angle street or right angle lane. My question is what is the maximum length of this ladder L that he can maneuver across this lane if this length is known to be A and this length is known to be B. So what I've drawn is basically a situation when, where he is in a deadlock position. That means his ladder is just about to cross or just about to get stuck. You can call this as a situation where it is just about to pass. That means he can now maneuver it in this way. I hope you can understand the situation. So he's standing here holding it like this and trying to take it this way. This person is Venkat. Stick diagram. Venkat stick diagram. So what is the maximum length of this ladder L that he can maneuver across this lane, which is right angular and this lane is of with A and this lane is of with B. This question can come even for your school exams. Of course, small change will be there in the question. They may not use the word Venkat there. So ladder is your objective function. So try to write ladder in terms of a thing which will keep on changing. In the interest of time, I'll also work with you because else I'll not be able to complete this chapter. So let's say this angle is theta. Okay, let's say, let's say this angle is theta. I'll just make a, okay, let's say this angle is theta. If this angle is theta, you would all admit that even this angle will be theta. Okay. So what is this length of the ladder? Let's say X and say this length of the ladder is Y. So can I say X will be A cos theta and Y will be B cos theta. This is your B. Okay. So total length of the ladder which is X plus Y is actually A cos theta plus B cos theta. Now, basically you have got an objective function which is in terms of a variable theta. Okay. So for your ladder length to be maximum or minimum, the derivative of L with respect to theta should be 0. Correct. Which means minus A cos theta cot theta plus B sec theta tan theta should be equal to 0. Correct. Which is nothing but saying A cos theta by sine theta should be equal to B sine theta by cos theta. In one minute. A, yeah. This will become sine theta by cos square theta. I'm sorry. And this will become sine square theta. Correct. In other words, in other words, I can say A by B. A by B will be tan cube theta. That means tan of theta is A to the power one-third B to the power one-third. Okay. So if tan theta is this, what is cos theta? Can I say cos theta would be under root of this square this square which is A to the power two by three B to the power two by three divided by A to the power one by three. Hypotenuse by perpendicular. So think in your mind this to be a perpendicular, this to be a base. So this is your hypotenuse and this is your perpendicular. Okay. Similarly, what will be your sec theta? Sec theta would be again hypotenuse by base. Now why I found out cosik and sec is because in the expression of the ladder I need to use those. Added. Cosik and sec. Yeah. So multiply this with an A. So your A cosik theta plus B sec theta would be A to the power two by three under root of this plus B to the power two by three under root of. Take common. So this will become A to the power two by three B to the power two by three whole to the power of three by two. So this will be your length of the ladder. Okay. Now I'm not doing the double derivative test to check whether this was the point of maxima and minima. So I'm leaving up to you to do that. So please, please do the double derivative test. Please check that d two L by d theta square should be negative for you to have the maximum length of the lab. Okay. Any questions here? Approach is clear how it works. So in this question, one learning is there is that when you are trying to find out the objective function, you can write it in terms of a assumed parameter. For example, theta was nowhere mentioned in the question. So I assumed theta to make my life easy. So don't shy away from taking this approach. Many people, they don't like taking this approach as a result, the problem becomes super lengthy. Let's take this question. There's a small change in the function. This will be for action. So read the question everyone. Let me draw the scenario for you all so that the understanding is clear. So there is a triangle ABC. Let's say ABC and on this triangle. Okay. They're given the coordinates of these points also. So is this B is this and see is this. Okay. Now on this triangle, they're choosing some points. For example, on BC, they're choosing D. So BC, they're choosing a point D. Let's say, okay. AC, they're choosing a point E. So let's say E is here. Okay. And F, they're choosing on AB, something like this. Now, by connecting these lines, they are creating AFD. AFD would be a parallelogram. That's what they're claiming. So these points are located in such a way that AB, AFD is a parallelogram. Using calculus show that the maximum area of this parallelogram is this expression. 1 fourth P plus Q, Q plus R, Q minus R. How would you solve this question? ITJ in 1986. So if this is a parallelogram, that means this is parallel to this. And this is parallel to this. Okay. Okay. So can I say, let's say this, this. Please wait. Okay. You're trying to solve it on your own. Okay. Okay. Take your time. Next class, I'll be starting with definite integrals. That's again a lengthy chapter. So three classes minimum will go. Okay. Let's, let's proceed as it'll be too much delayed. And by naming convention of a triangle, let's say I call this as B and I call this as C. So we all know in a triangle, the side opposite to an angle is named by the small alphabet for that angle. So this is side B. And this whole thing is side C. Can I say triangle CED, CED is similar to triangle CAB, right? Because this is parallel to this. This is a common angle. So these two angles will also be the corresponding angles. So by using my basic proportionality theorem, can I say, can I say that, this length by whole length will be this length by this whole length? So can I say EC by AC? EC by AC is ED by AB. Correct? Now what is EC? EC is B minus Y. AC is B. ED is X. AB is C. Correct? So can I say from here, 1 minus Y by B is equal to X by C. That means Y is equal to B times 1 minus X by C. Okay? Let's call it as 1. Now, we all know that, let's say this is angle A. We all know that area of this parallelogram, let me call this area of the parallelogram by P. So let P be the area of the parallelogram. We all know that area of the parallelogram in this situation will be XY sine A. Am I right? Now, A angle is fixed. So this is a constant term. X and Y are two variables. So as you can see, this is my objective function. This is my objective function, which is in two variables. Now I need to convert it to one variable, which I can easily do because of 1. So by using 1 or we can say from 1, I can convert it to a single variable function. So let X be X. Let Y be B times 1 minus X by C sine A. Now if you want this function to be maximum, its derivative with respect to X should be 0. Okay? Let's do the derivative of this. B sine A will be constant. So you have X minus X square by C derivative. This should be 0. That means B sine A, 1 minus 2X by C should be 0. In other words, 1 minus 2X by C should be 0. That means X should be C by 2. If X is C by 2, Y will be... What will be Y if X is C by 2? B by 2. Correct? D by 2. Right? That means the area of the parallelogram would be X, Y, that is B, C by 4, sine A. Right? Now we all know that it is actually half of half B, C, sine A. And what is half B, C, sine A? In any triangle, what does half B, C, sine A signify? Area of the triangle. Area of the triangle. Correct? So your answer would be your maximum area would be half of the area of the triangle A, B, C. Okay? Now in order to find the area of the triangle A, B, C, I will use its coordinates which has been specified to me in the question. So these are the coordinates. We all know how to find the area of a triangle when its coordinates are specified, which is half X1, Y1, 1, X2, Y2, 1, X3, Y3, 1. Correct? So it's P square minus P1, Q square Q1, R square minus R1. And you guys have all done your determinants. I'm sure you'll be able to solve this. So let us do this operation. So you'll do this operation R2 is R2 minus R1. And let's do R3 is R3 minus R1. So let us subtract the first row with all the other rows. So it's Q square minus P square, Q plus P0. R square minus P square, P minus R0. Okay? Let's take few things common. For example, I can take Q plus P common over here. And I can take P minus R common for me. So I'll get P square minus P1. And this will be Q minus P1, 0. This will be minus R plus P1, 0. Okay? So let us expand it now. Let us expand with respect to the third. Expand with respect to the third column. What does this give you? It gives you half P plus Q P minus R. This will give you Q minus P plus R plus P, which is actually Q plus R. Okay? So area of the triangle is given by this term. So area of the parallelogram would be again, half of this will be 1 fourth, 1 fourth P plus Q. Let me write Q plus R first. And let me write this as P minus R. I think this is what we wanted to prove, right? Check it out. Yeah. 1 fourth P plus Q, Q plus R, P minus R. Yeah. Okay. Is this clear? Let us take another one. By the way, anybody who wants to copy this, please do so. I think the idea is clear now what to do. Should I go to the next page? Yes. Some easier ones. Show that the semi vertical angle of a right cone of a given total surface area, including the area of the base and maximum volume is given by sine inverse one third. Done. So let's say the semi vertical angle is theta. Now, what is given? The total surface area is given. Total surface area will be pi RL plus pi R squared. Okay. This is given as a constant. Okay. Let's call it as C. Okay. So read the question. Clearly states that the surface area is given. Given means it is fixed. You can't change it. The volume is the objective function. So volume of a cone is one third pi R squared H. So this is your objective function. And this is your constraint. This is your constraint. Now your objective function is present in two variables, R and H both. Right. So how can we change this? H is R cot theta. Right. H as R cot theta sir. R cot theta. Okay. And where will I get theta from then? The semi vertical angle. Okay. So let's say R in terms of theta. Yeah. I take this as R. Okay. Now what is given to us is pi RL. Pi RL. Now this is your L. Okay. So what is L actually? In terms of R and theta. L is R by sine theta. Sine theta. Right. So pi R squared is given to us as a constant. Correct. Yes or no? In other words. In other words. Pi R squared which you can take as a constant. 1 by sine theta plus 1 is equal to C. Okay. So you can write pi R squared as C times sine theta by sine theta plus 1. What is H? What is H? H is R cot theta. Yes sir. Okay. So what do you do in this case? One minute sir. Yeah. Theta would make it complicated. Don't you think so? Because now you have to write your R square in terms of it. And then you have to square this and then you have to differentiate it. Why don't we take a non-theta approach? Sir, isn't there any way of scanning? Can't do that sir. Go ahead. It's actually your call. See what you have to do here is you have to write R squared in terms of theta which you can place over here. But H is also having R in it. Correct. So we have to square this up. Okay. So there are multiple ways to do it. So, okay, let's follow this approach only. So now, if you see this term, V square is one ninth of pi square R to the power four H squared. And H squared is equal to R squared cot squared theta. Correct. That means H squared is equal to R squared. R squared is this term, okay? By pi. This into cot squared theta. So basically you have one third. This term is the square of this term which is C squared sin theta by sin theta plus one the whole square. Into H squared. H squared is this term which is one by pi. Or you can say C by pi. This will become cube into cot squared theta. This will become your volume square. Now you must be wondering why am I interested in volume square is because if V is max, that means V square will also be max. So I can say D by D theta of V square will be equal to zero. So right now you need to, if you collect all the constants together, this will be your constants. And the variables would be sin cube theta by sin theta plus one whole cube into cos square theta by sin square theta. So here you can cancel off, make it sin theta and differentiate this with respect to theta and put it zero. Can we write sin theta as D and differentiate with respect to D because anyways we have to find out the sin theta. Oh yes sorry, this will be one ninth. Oh sorry, what did you say? Can we use sin theta itself as a variable? Yeah, why not? That's a very good idea. So Trippan has a very good idea here. He can, he's suggesting let's write sin theta as T. So now you are differentiating D, T1 minus T squared. By the way, a lot of terms will get cancelled here also. Actually it's not a very tedious job. So you have to do this. Yeah, correct Trippan. So we can write it as DT of, this is T1 minus T1 plus T that will get cancelled with this guy. So it will be 1 plus T the whole square. Into DT by D theta it should be equal to zero. DT by D theta is equal to zero. Let's forget about it. Let's differentiate this. Oh sorry. This will be 1 minus T square whole square. Derivative of this will be 1 minus 2T. Then minus T1 minus T into derivative of this will be 2 1 plus T. Okay. Denominator I'm not considering. I'm just saying it is equal to zero. And DT by D theta which is actually cos theta. DT by D theta is cos theta cannot be zero because if it is zero that means theta will be pi by 2. That means your semi-vertical angle is pi by 2. Means your cone will be almost like flat. Okay. So let's simplify this. Out of this a lot of things will be cancelled. For example, I can cancel off 1 plus T from all the sides because I know 1 plus T cannot be zero. So if you simplify this, you'll end up getting 1 minus 2T plus T which is minus T minus 2T square. Here also you'll get plus 2T square and minus 2T equal to zero. This and this gets cancelled off. So you end up getting 3T equal to 1. So T is equal to one-third. T equal to one-third means sin of theta is going to be one-third. Which means theta is equal to sin inverse one-third. Okay. See actually it's your call. How you want to solve it. But many people they get muddled up in the expression of this term because they're not able to convert it to a single variable. And this is a very good question. Please mark it with a star. If it comes in your school exam, most of you would waste a lot of time on this. I'm sure. Okay. So here you can write because DT by D theta cannot be zero. So only this is possible. After this you know how to proceed. So no need to copy after that. Can I move on to the next question? Find the points on this curve whose distance from the origin is minimum. This question actually can be solved without maxima minima also. I mean the the role of maxima minima is very less in this. Any idea how to do this? Anybody? Just give him. Okay. Okay. Take your time. No worries. Okay. Let's say the point is X, Y, which is at a minimum distance from the origin. So what did I do is I took this distance as R and the angle theta here so that I can write my point as R cos theta and R sin theta. So this is the point which I'm looking for whose distance is minimum. So first of all, this must satisfy this must satisfy the given curve. This must satisfy AX square to HX Y plus AY square. Secondly, I want R to be minimum. Correct. So R is my objective function and constraint is that R cos theta comma R sin theta must satisfy this curve. In other words, you're saying A R square cos square theta plus 2H R square sin theta cos theta plus A R square sin square theta is equal to C. If you take R square common from these two terms, you'll end up getting A and from these two terms, you'll end up getting H sin 2 theta. So this is going to be a C. So R square is equal to C by A plus H sin 2 theta. Now if your R is minimum, if your R is minimum, it implies your R square should also be minimum. Now, the reason why I said you do not need calculus for this is because if you want your R square to be minimum, this should be maximum. Correct. And this could be maximum only when sin 2 theta is equal to 1, which anyways calculus will also give you the same thing. But why do you use calculus when it can be done without it also? That means theta should be equal to 45 degrees. Correct. That means your minimum value of R would be under root of C by A plus H. But the question is they're asking us the point. Okay. So your point would be, your points would be, can I say it would be plus minus R cos theta comma R sin theta. So R cos theta comma R sin theta is under root of C plus A plus H twice. Remember sin 45 degree is 1 by root 2. So root 2 I'm introducing here. So this will be your points which will be at a minimum distance from the origin. Sir. Sir, I think I did it in a different method. But it's not nice actually. You got the answer. Sir, I did it in different ways. I took normal and all that. Okay. The basic principle which I used was it will lie on the normal to that curve also. Like whatever curve it is, it will lie on the normal also. Like first you differentiate that A x squared and all that. So the point which is at the minimum distance from the origin should be that point where the normal drawn should pass through the origin. Yes. And then you get some big expression. But eventually the answer is the same. Okay. Let's do the last question for this topic. After that, we are going to start with definite integral enough of AOD now. Next class problem solving. Next class problem solving. For a circle x square plus y square is equal to r square. Find the value of r for which the area enclosed by the tangents drawn from the point 6,8 to the circle and the chord of contact is maximum. Okay. So it's a case of a tangent drawn from an external point to a circle. So let's say. Okay. So this is your chord of contact AB and this is your RR. This is your origin. You want the area of a PB triangle to be maximum. Okay. Okay. So for this, we don't need a lot of coordinate geometry. We can stick to geometry actually. So let's say this is theta. Then can I say this angle will also be theta? Yep. So what is the base of the triangle? What is the base of the triangle? Can I say base of the triangle will be 2r cos theta? Yes. Correct. What will be the height of the triangle? We all know that this is going to be 90 degrees. So height of the triangle, let me call it as PM. Can I say height of the triangle will be 10 minus r sin theta? Why 10? Because O2P is 10 and this guy is r sin theta. So area of the triangle, ABB will be half base into height. So half into 2r cos theta into 10 minus r sin theta. So 2 2 gone. Okay. Is this fine? Any concerns with this? No sir. Okay. Now can I say r is also given by 10 sin theta? Because this is 10. So I can write 10 sin theta for this. So 10 sin theta cos theta 10 minus r sin theta. Here also I can replace. Okay. So this is your area of the triangle. Let me call it as A for the time being. Good enough. Okay. Now the problem is we have to differentiate this. Yeah. So this will be 10 square which is 100 sin theta. Sin square will be cos cube theta. Correct. So this is your A. So dA by d theta would be equal to 0. Let's see what happens. 100 sin theta derivative will be cos theta. So I can write it as cos to the power 4 theta minus 3 cos square theta sin square theta. This will be equal to 0. Correct. From here can I say cos to the power 4 theta will be 3 cos square theta sin square theta. In other words tan square theta will be 1 by 3. So tan theta is equal to 1 by root 3. That means theta is equal to 30 degree. Correct. So your minimum value, your r value should be 10 sin theta. 10 sin theta is 10 into sin of 30 degree which is equal to 10 into half. Answer is 5. How can we solve this with conics? So first of all we know that, I'm not sure actually, I was doing it but then I lost out on it. It was taken too much time. Okay. So first we know that A B is going to be 6 x plus 8 y plus r. Equal to r. No, no, no. Okay. This is things like I derived that now. Area of that triangle is going to be rl cube by r square plus l square. What is l? The length of the tangent sorry. Okay. And then I tried differentiating but it didn't work sir. It didn't work. I think it was taking too long. Yeah, it will take too long. That's very obvious. What geometry can do here, coordinate geometry will not be able to do. It's a simple case of a geometry. Okay. Okay. So fine, we'll stop here. And next class, please do not miss it because it's going to be a very important class. We'll be talking about definite integrals. Okay. Thank you. Bye bye. Take care. Have a great day. Bye sir. Bye sir. Have a good night. Bye.