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
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.
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?
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.
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.
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.
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 .....
that is one of the cleanest of 14.8 I have seen, using textbook type solving techniques. ty.
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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
الله يسعدك يارجل ماتوقعت ان الموضوع بسيط للدرجة هذه😮😮
Thanks you so much! Saved my efforts from scratching textbooks😀
That's a great lecture! Thank you so much for your time and knowledge, sir!
Example 2 was basically identical to one that was driving me crazy- couldn’t figure out. Thanks for the help!
same
Thank you so much. I have an exam tomorrow and this helped me a lot.
saved my day. you are the man.
This man is a legend
loved it. saved the day!!
I needed this.
Thanks for this video. Not enough CZcams videos on Calc 3 :)
This was very helpful. Thanks
thx lol you make it clear for me
thankyou❤️It helped me alot❤️
Great video
Thank you, great video for practise
thank you very much for making video in detail
thanks a lot!
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You are my savior
Thanks you helped me alot
GOAT
Thank u man, thank u so much
If subject to is x+y=0 ,how do we have to put it??
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.
Amazing
Masterpiece
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.
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.
@@HamblinMath thank you :)
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?
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.
5x+6y-b=0=g(x,y) which is your constraint. f(x,y)=xy is your objective function. So yes you were correct
the best
Hello, for question 2- why did you differentiate -4x^2 for the f(x) value? I thought we only differentiate g(x,y)? Thanks
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
@@HamblinMath Many thanks
Slay!
So helpful thank you so much indeed
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?
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.
@@HamblinMath thanks!
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.
How did you minimize the root?
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.
@@HamblinMath Got it. Thanks!
4:21 i lost it from here
why question 2 the lamba 1/2 ignored?
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 .....