Lagrange Multipliers Practice Problems

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Lagrange Multipliers Practice Problems

Рет қаралды 137,695

James Hamblin

Күн бұрын

Пікірлер: 54

@N0N5T0P 6 жыл бұрын

Not all heroes wear capes.

@torlarsen2212 4 жыл бұрын

nonstop how do you know he’s not wearing a cape?

@utkarshsrivastava2326 2 жыл бұрын

But he is wearing cap u can see in his you tube account display picture

@xinfap.5968 5 жыл бұрын

that is one of the cleanest of 14.8 I have seen, using textbook type solving techniques. ty.

@TrueArmenianBoss1234 6 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

@rmb706 4 жыл бұрын

Example 2 was basically identical to one that was driving me crazy- couldn’t figure out. Thanks for the help!

@adityaparanjape3827 8 ай бұрын

same

@Salamanca-joro 4 ай бұрын

الله يسعدك يارجل ماتوقعت ان الموضوع بسيط للدرجة هذه😮😮

@surbhi57866 5 жыл бұрын

Thanks you so much! Saved my efforts from scratching textbooks😀

@meghanath2171 3 жыл бұрын

Thank you so much. I have an exam tomorrow and this helped me a lot.

@pedrocolangelo5844 10 ай бұрын

That's a great lecture! Thank you so much for your time and knowledge, sir!

@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.

@daltonjberkley44 5 жыл бұрын

This man is a legend

@kavinyker6837 3 жыл бұрын

saved my day. you are the man.

@isaachossain2807 Жыл бұрын

I needed this.

@Iusedtobescene 2 жыл бұрын

Thanks for this video. Not enough KZfaq videos on Calc 3 :)

@danielj5650 4 жыл бұрын

Was looking for videos on the song la grange and ended up here

@Darth_Cassius_ 4 ай бұрын

Thank you, great video for practise

@pratikwaghmode7311 4 жыл бұрын

thank you very much for making video in detail

@vidwanshisood3227 5 жыл бұрын

thankyou❤️It helped me alot❤️

@sadeq__kh 2 ай бұрын

Amazing

@RedBanana44 4 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?

@Emeryx 3 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.

@eduardomoreira7624 3 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

@ralphmichael3355 6 жыл бұрын

loved it. saved the day!!

@steveying1305 4 ай бұрын

GOAT

@mohammedshalabi4191 2 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.

@poetryaddict1 6 жыл бұрын

This was very helpful. Thanks

@gp7493 4 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.

@HamblinMath 4 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.

@gp7493 4 жыл бұрын

@@HamblinMath thank you :)

@asadzaman5573 4 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

@HamblinMath 4 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

@asadzaman5573 4 жыл бұрын

@@HamblinMath Many thanks

@alecchristophergossai7956 4 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?

@HamblinMath 4 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.

@alecchristophergossai7956 4 жыл бұрын

@@HamblinMath thanks!

@GiZm0865 5 жыл бұрын

You are my savior

@gumoshabeclaire2762 3 жыл бұрын

Thanks you helped me alot

@santiagoreyes9440 3 жыл бұрын

Great video

@JMac___ Жыл бұрын

Thank u man, thank u so much

@abdullahaljhani9754 5 жыл бұрын

thx lol you make it clear for me

@MrAbbasalrassam 6 жыл бұрын

So helpful thank you so much indeed

@user-tw3mk6dv8o 8 ай бұрын

Masterpiece

@Salamanca-joro 4 ай бұрын

4:21 i lost it from here

@rohitahijam813 5 жыл бұрын

If subject to is x+y=0 ,how do we have to put it??

@shehryarmalik5704 6 жыл бұрын

thanks a lot!

@assil110 5 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 .....

@lesliesie3506 3 жыл бұрын

why question 2 the lamba 1/2 ignored?

@annas7853 10 ай бұрын

Slay!

@trm_tba9820 4 жыл бұрын

the best

@abdullahhaider4833 5 жыл бұрын

How did you minimize the root?

@HamblinMath 5 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.

@abdullahhaider4833 5 жыл бұрын

@@HamblinMath Got it. Thanks!