You don't need calculus for this question

[@blackpenredpen]

How Cauchy would find the maximum of sqrt(x)+sqrt(y) kzbin.info/www/bejne/j4Stq6OHhriCkNk

[@aks8403]

Hello sir, i tried to apply the RMS>AM inequality in this question you solved with Cauchy inequality, but i didn't get the correct result from it, could you help figure it out?

[@ShaunakDesaiPiano]

If you don’t know or think of the AM-GM inequality, you can instead differentiate the expression with respect to xy to get the same minimum.

[@Undefined_infinity]

He has said without using calculus blud.

[@ShaunakDesaiPiano]

@@Undefined_infinity oh I read the thumbnail instead of the title, my mistake.

[@ianfowler9340]

Or you could start with: For a,b >=0 then it must be true that: ( √a - √b ) ^2 >= 0 a - 2√a√b + b >= 0 a + b >= 2√(ab) Same thing really, just a little shorter. My high school Algebra teacher used to jokingly call these "magic line" proofs. As in " how did you know ahead of time " to pick that particular true statement to start with in order to arrive at the true statement you want ? LOL. Good video.

[@blackpenredpen]

@@ianfowler9340 ah yes. Thank you!

[@avinotion]

I was about to suggest the same thing. Once you've done enough integrals, those square formulas start popping out instantly (which makes it really easy to remember the proof for the roots of quadratic equation) - even after over 10 years of not dealing with math.

[@Songfugel]

Thanks for the notes at the end! I was feeling a bit let down by the answer without that part 🤘

[@AyberkDurgut]

Inequalities are so important.

[@blackpenredpen]

Definitely!

[@GhostyOcean]

They're fundamental to analysis!

[@anglaismoyen]

Makes sense. There are so many more ways things can be unequal than equal.

[@AmmoGus1]

Especially between people! Equality is a myth, a complete fantasy. We will only kill ourselves in trying to achieve it. Sad!

[@janami-dharmam]

but some are more equal than others!

[@mcalkis5771]

A video explaining Lagrange multipliers would be most welcome.

[@boredomgotmehere]

This is why you’re in my top 3 of YT maths tutor. Google should have an award category for maths teachers on YT where we can vote for the best that created the most value for math lovers. ❤️ and cheers from me. Hope you continue stuff like this.

[@blackpenredpen]

Thank you.

[@RadChromeDude]

oooooo nice one! i'm taking an optimisation module this sem - Lagrange multipliers and KKT points are such elegant things :)

[@abbyskywind]

another solution, algebra but not using the AM/GM inequality directly: let F = xy + 16/xy + 10 let G = xy + 16/xy, observing that F is minimal when G is minimal then let u = xy, so we have G = u + 16/u G*u = 16 + u^2 ( i ) u^2 - G*u + 16 = 0 by completing the square we see that this is minimal when u = G/2 so from ( i ) we have: G*(G/2) = 16 + (G^2)/4 G^2 = 64 G = 8 since G > 0 hence the minimum = F = G + 10 = 10 + 8 = 18

[@ExplosiveBrohoof]

My immediate thought before watching the rest: It expands to xy+10+16/(xy). Now let u=xy and use calc 1.

[@tommychau1211]

First glance, cauchy schwarz inequality 1. sqrt x & sqrt y are legit and checked (as both >0) 2. Direction of inequality is correct (x+2/y)(y+8/x) = [ (sqrt x)^2 + (sqrt 2/y)^2 ] [ (sqrt 8/x)^2 + (sqrt y)^2 ] >= ( sqrt x * sqrt 8/x + sqrt 2/y * sqrt y)^2 = (sqrt 8 + sqrt 2)^2 = (3 *sqrt 2)^2 = 18

[@Genzoo7]

We can substitute t=xy, find d/dt f(t), assume xy=4 -> min, so f(1,4)=f(2,2)=f(4,1) = 18

[@BPEREZRobertJamesL]

I thought of the Cauchy-Schwarz Inequality immediately after seeing the problem. This makes it trivial: (x + 2/y)(y + 8/x) = ((√x)² + (√(2/y))²)((√y)² + (√(8/x))²) ≥ ((√x)(√(8/x))+(√(2/y))(√y))² = (√8 + √2)² = (3√2)² = 18 with equality when (√x,√(2/y)) and (√(8/x),√y) are linearly dependent, which happens if and only if xy = 4 (the proof is left as an exercise to the reader). So, 18 is the minimum of the expression.

[@bartekabuz855]

Now prove AM-GM for any n without Lagrange multiplier

[@dikshantgoyal8270]

Use jensens inequality for lnx

[@MA-bm9jz]

​@@dikshantgoyal8270prove ln is concave without derivatives

[@bartekabuz855]

@@dikshantgoyal8270 That doesn't simply things XD

[@dalek1099]

I think I did it Analysis I without knowing about Lagrange Multipliers. I didn't know about Lagrange Multipliers at the time. I did it by induction proving n+1 implies n and then n implies 2n, which leads to n implies n+1. It was quite a tricky homework question.

[@MA-bm9jz]

​@@dikshantgoyal8270 now prove ln is concave without derivatives

[@dan-florinchereches4892]

Wow this use of the means inequality rocks. When i saw xy+16/xy i tought about the minimum of a^2+16/a^2=a^2+16/a^2+8-8=(a-4/a)^2+8=>8

[@ausaramun]

I used the fact that the intersection point of xy and 16/xy give you a minimum sum through the use of graphs.

[@anonymouscheesepie3768]

Let a = xy Our function is M(a) = 10 + a + 16/a, a > 0 To find minimum, differentiate M with respect to a dM/da = 1 - 16/a^2 At a minimum or maximum, dM/da = 0 1 = 16/a^2 16 = a^2 a = 4 (since a cannot be negative) At a = 1, dM/da is negative At a = 5, dM/da is positive Thus a achieves a minimum at a = 4 If a is 4, then our minimum M(a) = 4 + 10 + 4 = 18

[@absxn]

Wouldn’t it be easier to differentiate the expanded expression with respect to xy, then set = 0? This gives xy=4 which gives the same answer when you plug back in

[@tomekk.1889]

He just wanted to show off this specific way of solving the equation

[@Musterkartoffel]

Thats at 4:45 😅

[@tomekk.1889]

@@Musterkartoffel Man don't judge us nobody watched that far before commenting 😂

[@absxn]

@@Musterkartoffel hahahaha u right u right

[@BajoJajoBajoJajo2]

He said ini de beninging he wanted to do it without calculus, and only algebra.

[@carlosangulo2888]

Lagrange is usted in non linear programing to find solution to non linear problems in optimization.

[@hebertysouza5671]

Maybe multiply everyone by xy and make a quadratic equation and find the value of m (min) that makes delta = 0. x²y²+10xy+16-m=0 0 = 10²-4*1*(16-m) 0 = 100-64+4m 0 = 36 + 4m m= -36/4 = -9 But m has to be a number bigger than 10, so i has make something wrong. I see the mistake was by define the expexion equals m after multiplied by yx, try again in the right way, i just will leave this mistake here because yes. xy+10+16/yx = m x²y²+10xy+16=m*xy x²y²+(10-m)xy+16=0 Now, delta equals 0 0 = (10-m)²-4(1)*(16) 64 = (10-m)² 10-m = ±√64 10±8=m m = 2 : : This can not be m = 18 Lets see if i make this right

[@charlottesphie7037]

My sensei always use the AM GM method for maximum volume. Idk how he do that but its very effective.

[@ChinalurumUkairo]

What an amazing video bprp! I am only an 8th grader but ever since I saw your calculus videos, I have been engrossed with your channel. Is it ok if you can make a video with a question featuring an improper integral, a derivative, a series and a limit all in one question where it is aimed at Calculus 1 and 2 students. Thank you!

[@datbubby]

You could also just set gradient == 0 x^2 * y^2 == 16, x,y > 0 xy == 4 when gradient == 0 xy + 10 + 16/xy == 4 + 10 + 16/4 == 18

[@abtinhidaji9977]

I solved it in another way, substituting t=xy and taking the derivative equal to zero, resulting in t=4 therefore xy=4 and the minimum is 18.

[@ianfowler9340]

BTW, a more general method for finding stationary points - but a calculus one - for a function of 2 variables. For f(x,y) = (x + 2/y)(y + 8/x) , find the 2 partials (pf/px and pf/py) and set them both = 0. This will result in a system of 2 equations in the 2 variables x and y. Solve this system and you will obtain y = 4/x or y = -4/x. This will, in turn, lead to a max value of 18 and a min value of 2. Lot more algebra and a lot more time. But, in this example, we have the LUXURY of being able to express f as a single function of t by the substitution t = xy - not always the case in general for functions of 2 variables. Having said that, I must stress that this video shows us the power of how approaching a problem from a completely different perspective can not only sometimes simplify the solution, but also lead to some valuable insights about the behaviour of the function itself - that we may otherwise have missed. Well done.

[@_Exen_]

i’ve been trying to integrate sin(6x)cos(x) for so long and i already tried every single integration technique but i just can’t figure it out

[@JoaoVitorSilveira-yd8gi]

Have you tried transform this product in a sum ? You can use sinP + sinQ = 2sin(P+Q/2)Cos(P-Q/2)

[@_Exen_]

@@JoaoVitorSilveira-yd8gi you say expanding the integral?

[@JoaoVitorSilveira-yd8gi]

@@_Exen_ yeah, you can turn this into two integrals using this formula, i guess thats the weiertrauss formula or something like that. Not sure If Works, i havent tried to solve It yet.

[@_Exen_]

@@JoaoVitorSilveira-yd8gi when i tried expanding i got like 10 terms multipling in order to make the sin and cos arguments just “x”

[@JoaoVitorSilveira-yd8gi]

@@_Exen_ Nah bro, you dont need that. sin(6x)cosx = 1/2(sin7x + sin5x)

[@Hussain-px3fc]

How do you proof the general case for the AM-GM inequality where you have n-terms?

[@ChessEDITZ2]

Here is a challenge for you BPRP:- There are 'n' terms in an A.P. and n-1 terms in a G.P. The common ratio is equal to the first term of the A.P. The (n-1)th term of the A.P. is a multiple of the common ratio. The nth term of the A.P. is 100 more than the 2nd term of the G.P. The A.P. and G.P. only include whole numbers. The sum of the last terms of both the progressions is 100. The first term of the G.P. is equal to 0.28 more than the co-ordinates of the foot of the perpendicular from (-1,3) to the line 3x-4y-16=0. Find both the progressions ,if the total number of terms in G.P. does not exceed 5 and the last term of the A.P. does not exceed 30.

[@joe40173]

AM-GM不等式的證明過程中並非總是iff的喔~ 這也是AM-GM要求每個items都要大於0的原因

[@ThijquintNL]

Just found out i haveng been subscribed after whatching for years, so do more plugs. Anyway i had my first ever calculas exam ever in my inversity and so thank you for the years of prep work :) Also I could only get x=4 on (2x²-3x)/(x+1)=4 and photomath's method seems very subjective. How do you get the x=-0.5 solution?

[@amiralidorodian1932]

can you please do a tutorial on Green's theorem on your Calculus Channel?

[@GianMichelAbreuDelgado]

I loved your video ❤

[@Caravanseraii]

Can we use cauchy shwartz?

[@YuukiKobayashi-wg9mr]

Hello there, just tried it Great question, looking forward to more math questions

[@armanavagyan1876]

Thanks PROF

[@nayeem7359]

Wow, learnt a nice inequality today thanks to you 😇

[@Himanshusingh.123]

f(x)=Integrate[e^|x-t|,{t,0,2}] ; find mimimum of this function Pls do this

[@Ninja20704]

We can also use cauchy schrawtz to do it but the setup would be a bit tricker compared to AM GM

[@varadthube2396]

For cauchy inequality setup is much easier...... Dont even need to expand...

[@Ninja20704]

@@varadthube2396 I don’t mean it in that way. In order to use CS you need to write each of the individual terms as (…)^2 so for example you would need to write x as [sqrt(x)]^2, 2/y as [sqrt(2)/sqrt(y)]^2 etc before you can apply the inequality, which some ppl find slightly unobvious/unintuitive. That was what I meant by “tricker”

[@tvvt005]

4:47 why is the min when Xy=16/(xy)

[@hlib_tarasenko]

This follows from AM=GM if and only if a=b. Which is further explained in the video. Hope this helps

[@lelouch1722]

We could have optimized a 1D function by using t=xy and then looking at (x,y) tuple corresponding to optimal t=xy

[@alexicon2006]

Never heard of Lagrange Multiplier as we have a different High School Curriculum in India, but the Mean Inequality might actually be one of the most goated and clutch rules of maths Ive learnt so far. It. Is. So. Versatile. I swear to god.

[@carultch]

The idea of Lagrange multipliers, is constrained optimization in multivariable calculus. The underlying idea is that you have a constraint equation between x & y, and you have z equal to a function of x & y. The local maximum or local minimum of z, will occur when the contour lines of the function of z, are parallel to the constraint equation between x & y. Consider a hill defined by: z = 1/4*(9*x^2 + 4*y^2 + 1)*e^(-x^2 - y^2) And consider a trail along this hill, that follows the ellipse: 4*x^2 + 9*y^2 = 9 We aren't interested in extremes of the hill alone. We want to know, where along the elliptical trail, the height is maximized. To do so, we take the gradient of the hill, and the gradient of the ellipse, and then equate them to each other, with a scale factor of lambda. Lambda is any arbitrary real number (other than a trivial zero), which is called the Lagrange multiplier. The gradient is a vector built of partial derivatives, that tells you the direction of greatest ascent, and slope of greatest ascent, at each point on the hill. It's common that you don't care to find lambda; just that it's consistent among the equations. This sets up the following system of equations to solve: x-direction partial derivatives: -1/4*x*(9*x^2 + 4*y^2 - 8)*exp(-x^2-y^2) = lambda*8*x y-direction partial derivatives: -1/4*y*(9*x^2 + 4*y^2 - 3)*exp(-x^2-y^2) = lambda*18*y Constraint ellipse itself: 9 = 4*x^2 + 9*y^2 Observe that there are four trivial solutions. Two when x=0 and y=+/-1, and two when y=0 and x=+/-1.5. As it turns out, these are all local minima along the trail. The non-trivial solutions give us the local maxima. Observe also, that after eliminating the trivial solutions, all instances of both x and y are squared. Let U=x^2, and let V=y^2. -1/4*(9*U + 4*V - 8)*exp(-U - V) = 8*lambda -1/4*(9*U + 4*V - 3)*exp(-U - V) = 18*lambda 9 = 4*U + 9*V The solutions for U & V: U = 72/65, V = 33/65 And all possible square roots of both U & V, will form the four corresponding solutions for x & y: x = +/- 6*sqrt(2/65) y = +/- sqrt(33/65)

[@Caravanseraii]

​@@carultchHe said he's a high schooler bro, do you expect him to understand all that?

[@GauthamRaajesh]

Well he didn't specifically state hes in hs, just that his hs didn't teach Lagrange multipliers. But yea lol, that's a bit hard to understand for hsers

[@alexicon2006]

@@GauthamRaajesh I just graduated from school grade 12th this year. So I have just finished school. Our Mathematics Syllabus in 11th and 12th involves Coordinate Geometry Calculus Sequence Series Sets, Relations and Functions Complex Numbers Combinatorics Statistics Probability Matrices and Determinants 3D Geometry and Vectors And I think thats about it. In coordinate we have the basics, straight lines, and all four conic sections. Calc is also pretry typical with limits then continuity then differentiability, followed by differentiation then its applications, and then indefinite and definite integrals abd their applications, and finally, differential equations.

[@GauthamRaajesh]

@@alexicon2006 ahh makes sense, where I live, we mostly learn that too, but sometimes, advanced students have the option to skip parts and jump to multivar their senior year. But mostly all students here take basically all of what you said.

[@aadityakiran07]

prove that 1!*2!*3!*4!..400!/200! is a square number

[@udk932]

You could just use school math: z^2*f'(z)=z^2-16=0 where z=xy>0 so z=sqrt(16)=4 and f=18

[@cybersolo]

I paused the video and tried this: Setting z = (x+2/y) * (y+8/x) and computing the derivatives dz/dx and dz/dy (they give the same result) we get y^2*x^2 - 16 = 0. ==> (y*x - 4)*(y*x + 4) = 0 => x = -4/y not valid because x,y > 0 So x = 4/y Replacing eliminates the variables and we found z=18.

[@taflo1981]

Just to clarify: The partial derivatives are not identical, but both dz/dx=0 and dz/dy=0 are equivalent to x²y²=16. Anyway, the derivatives being zero is just a necessary condition for having a minimum. One needs to prove seperately that these points are indeed minima. (Hesse matrix doesn't work, as it's semidefinite.)

[@mohamed2744]

But how will we solve it using Lagrange multiplier there's no constraint

[@carultch]

You wouldn't really need to think about Lagrange multipliers here, because just using the standard method of searching for stationary points, will find a continuous stationary curve of y=4/x. The partial derivatives both equal zero, at every point along this entire curve, so it is like a completely flat canyon on the 3D plot. Following this curve, you stay at a uniform vertical position of z=18.

[@antagonist4823]

This is what you learn at high school in the first year 😅

[@saigow349]

Hey. Can somebody help me with this integral? x^2/(x^4+1)

[@carultch]

Given: integral x^2/(x^4 + 1) dx Construct the quartic as a product of two arbitrary quadratics: x^4 + 1 = (x^2 + b*x + c)*(x^2 + d*x + f) Expand: x^4 + (b + d)*x^3 + (c + f + b*d)*x^2 + (b*f + c*d)*x + c*f Equate like coefficients: (b + d) = 0 c + f + b*d = 0 b*f + c*d = 0 c*f = 1 Solution: b = sqrt(2), c = 1, d=-sqrt(2), f=1 Let K = sqrt(2)/2, to simplify our writing Thus: (x^4 + 1) = (x^2 + 2*K*x + 1)*(x^2 - 2*K*x + 1) Complete the square on both quadratics: x^2 + K*x + 1 = (x + K)^2 + 1/2 x^2 - K*x + 1 = (x - K)^2 + 1/2 Thus: (x^4 + 1) = ((x + K)^2 + 1/2)*((x - K)^2 + 1/2) Set up partial fractions: x^2/[((x + K)^2 + 1/2)*((x - K)^2 + 1/2)] = (A*(x + K) + B)/((x + K)^2 + 1/2) + (C*(x - K) + D)/((x - K)^2 + 1/2) Plug in strategic values of x, to solve for A, B, C, and D. Using x=0, x=1, x=K, and x=-K, and then solving the system of equations, I get the following solutions: A = -K/4 B = 1/4 C = K/4 D = 1/4 Thus, our integrand becomes: (-K/4*(x + K) + 1/4)/((x + K)^2 + 1/2) + (K/4*(x - K) + 1/4)/((x - K)^2 + 1/2) The first term integrates with u-substitution, by letting u = (x + K)^2, and du = 2*(x + K). The third term integrates in a similar way, which we'll assign w=(x - K)^2 and dw = 2*(x - K). integral -K/4*(x + K)/((x + K)^2 + 1/2) dx = -K/8*integral u/(u^2 + 1/2) dx = -K/8*ln(|u^2 + 1/2|) = -K/8*ln(|(x + K)^2 + 1/2|) integral K/4*(x - K)/((x - K)^2 + 1/2) dx = K/8*integral w/(w^2 + 1/2) dx = K/8*ln(|w^2 + 1/2|) = K/8*ln(|(x - K)^2 + 1/2|) The 2nd and 4th terms integrate as arctangent. 1/4*integral 1/((x + K)^2 + 1/2) dx = arctan(sqrt(2)*(x + K)))/(2*sqrt(2)) 1/4*integral 1/((x - K)^2 + 1/2) dx = arctan(sqrt(2)*(x - K)))/(2*sqrt(2)) Put it all together: K/8*ln(|((x - K)^2 + 1/2)/((x + K)^2 + 1/2)|) + arctan(sqrt(2)*(x + K)))/(2*sqrt(2)) + arctan(sqrt(2)*(x - K)))/(2*sqrt(2)) + C Substitute K, factor out a master constant, simplify the ratio inside the logarithm, add the constant, and we're complete: sqrt(2)/8 * [ln(|((2*x - sqrt(2))^2 + 2)/((2*x + sqrt(2))^2 + 2)|) + 2*arctan(sqrt(2)*x + 1) + 2*arctan(sqrt(2)*x - 1)] + C Note that this C, has nothing to do with the C we used earlier.

[@saigow349]

@@carultch What?!! this is so much bigger than expected. Thanks for solving it for me😃

[@saigow349]

@@carultch I think I found a shorter solution drive.google.com/file/d/1xx6IrZJ82wFOq1fZI6WWEKXogyKhjIx6/view?usp=drive_link

[@saigow349]

@@carultch Hey, I found a much shorter solution. drive.google.com/file/d/1xx6IrZJ82wFOq1fZI6WWEKXogyKhjIx6/view?usp=sharing

[@saigow349]

​@@carultch Hey, I found a shorter solution. drive.google.com/file/d/1xx6IrZJ82wFOq1fZI6WWEKXogyKhjIx6/view?usp=sharing

[@ordinaryoreo8297]

didnt know how to do this so I just kinda thought maybe half of the numerator in the fraction so y=2/2 which is 1 and x=8/2 which is 4 and it turned out to be 18 so.. lucky guess 😅

[@basiliszag]

So clever

[@scottleung9587]

Cool!

[@nrxtfwd]

Cool

[@sheikhalsumaiya7230]

❤

[@ymj5161]

均值不等式秒了

[@Cool_Bungle]

...fo' this question

[@cdkw2]

I love abusing inequalities!

[@AmanYadav-id5gt]

You fogot the 'r'.

[@mzahmohamed6225]

M F the formule iFF

[@aMartianSpy]

$5

[@siddhantprasad1551]

Hello

[@blackpenredpen]

Hiii!

[@siddhantprasad1551]

Hello sir l am from class 10 and saw my big brother watching your video but still . I like watching your video I feel even if I am not getting ever thing you say but I learn something ever time I watch your video

[@lailalyyday]

helloo!!

[@blackpenredpen]

Hello!!