But he is wearing cap u can see in his you tube account display picture
@xinfap.59685 жыл бұрын
that is one of the cleanest of 14.8 I have seen, using textbook type solving techniques. ty.
@TrueArmenianBoss12346 жыл бұрын
Thank you so much sir, you have really helped me with the algebraic techniques. I don't know why, but Lagrange Multipliers has been by far the hardest calculus topic I've ever come across. The set up is easy, but the algebra is a nightmare
@Salamanca-joro3 ай бұрын
الله يسعدك يارجل ماتوقعت ان الموضوع بسيط للدرجة هذه😮😮
@rmb7064 жыл бұрын
Example 2 was basically identical to one that was driving me crazy- couldn’t figure out. Thanks for the help!
@adityaparanjape38278 ай бұрын
same
@surbhi578665 жыл бұрын
Thanks you so much! Saved my efforts from scratching textbooks😀
@meghanath21713 жыл бұрын
Thank you so much. I have an exam tomorrow and this helped me a lot.
@pedrocolangelo584410 ай бұрын
That's a great lecture! Thank you so much for your time and knowledge, sir!
@daltonjberkley445 жыл бұрын
This man is a legend
@kavinyker68373 жыл бұрын
saved my day. you are the man.
@Iusedtobescene2 жыл бұрын
Thanks for this video. Not enough KZbin videos on Calc 3 :)
@isaachossain2807 Жыл бұрын
I needed this.
@Darth_Cassius_4 ай бұрын
Thank you, great video for practise
@sadeq__kh2 ай бұрын
Amazing
@danielj56504 жыл бұрын
Was looking for videos on the song la grange and ended up here
@pratikwaghmode73114 жыл бұрын
thank you very much for making video in detail
@DaBestAround Жыл бұрын
Hey guys at 1:38, I would advise on not finding x and y individually like James has done in this example. The reason is that in other questions (such as example 3), solving the question via this method will be too cumbersome and it's not a method that can be extended to more difficult problems. The reason it looks so simple at 1:38 is that the example is really simple. Instead find two equations where you get lambda on its own. Once you have these two equations, equate them to each other. Once you equate these equations, after cancelling out some terms, you will get an equation for x in terms of y OR y in terms of x. Once you have this specific equation, substitute it back into the objective function and the question is pretty much solved.
@ralphmichael33556 жыл бұрын
loved it. saved the day!!
@vidwanshisood32275 жыл бұрын
thankyou❤️It helped me alot❤️
@poetryaddict16 жыл бұрын
This was very helpful. Thanks
@JMac___ Жыл бұрын
Thank u man, thank u so much
@gumoshabeclaire27623 жыл бұрын
Thanks you helped me alot
@MrAbbasalrassam6 жыл бұрын
So helpful thank you so much indeed
@steveying13054 ай бұрын
GOAT
@abdullahaljhani97545 жыл бұрын
thx lol you make it clear for me
@mohammedshalabi41912 жыл бұрын
Can you help me about this question Find the point (x, y) with the largest y value lying on the curve whose equation is y2 = x − 2x2 y.
@santiagoreyes94403 жыл бұрын
Great video
@RedBanana444 жыл бұрын
HI, the question I have is 'find the maximum value of xy subject to 5x+6y=b, where b is a positive constant. Does this mean f(x,y) = xy?
@Emeryx3 жыл бұрын
No, it doesn't! Since the partial derivative of your constraint (5x+6y - b = 0 is x + y) So that means your Lagrange function is L = f(x,y) + lambda(5x+6y-b) and then you go from there partial derivating for x and y. Then using the multiplier rate to find your max and min.
@eduardomoreira76243 жыл бұрын
5x+6y-b=0=g(x,y) which is your constraint. f(x,y)=xy is your objective function. So yes you were correct
@GiZm08655 жыл бұрын
You are my savior
@shehryarmalik57045 жыл бұрын
thanks a lot!
@gp74934 жыл бұрын
At 6:31, how did you decide that since the Greek letter is equal to -4 y has to be =0? A bit confused on that.
@HamblinMath4 жыл бұрын
We know that either y=0 or lambda=1/2. If lambda equals -4, then we know it *doesn't* equal 1/2, so y must be 0.
@gp74934 жыл бұрын
@@HamblinMath thank you :)
@user-tw3mk6dv8o8 ай бұрын
Masterpiece
@Salamanca-joro3 ай бұрын
4:21 i lost it from here
@asadzaman55734 жыл бұрын
Hello, for question 2- why did you differentiate -4x^2 for the f(x) value? I thought we only differentiate g(x,y)? Thanks
@HamblinMath4 жыл бұрын
Lagrange multipliers requires f_x = lambda g_x and f_y = lambda g_y, so you need the partial derivatives of both f and g
@asadzaman55734 жыл бұрын
@@HamblinMath Many thanks
@rohitahijam8135 жыл бұрын
If subject to is x+y=0 ,how do we have to put it??
@alecchristophergossai79564 жыл бұрын
for question 2, how did you automatically know that we can't solve for the Lagrange multiplier, and set them equal to each other (and then solve for y in terms of x and plug into original constraint). how will i know on a test to solve it your way?
@HamblinMath4 жыл бұрын
You can solve for lambda, but you'd have to divide both sides of those equations by x (or y). So you'll still have the case where x (or y) equals zero.
@alecchristophergossai79564 жыл бұрын
@@HamblinMath thanks!
@trm_tba98204 жыл бұрын
the best
@annas78539 ай бұрын
Slay!
@lesliesie35063 жыл бұрын
why question 2 the lamba 1/2 ignored?
@abdullahhaider48335 жыл бұрын
How did you minimize the root?
@HamblinMath5 жыл бұрын
Since sqrt(x) is a strictly increasing function, it is minimized/maximized exactly when x is minimized/maximized. It's a common trick that is used to simplify the derivatives in the case where we are optimizing distance.
@abdullahhaider48335 жыл бұрын
@@HamblinMath Got it. Thanks!
@assil1105 жыл бұрын
Nice video. Though, theoretically, we should calculate the determinant of the Hessian matrix to know whether the critical point is max/min/saddle point/or .....