Finishing the intro lagrange multiplier example

Working out the algebra for the final solution the the example from the previous two videos.

[collinsmith]

this was the BIGGEST help for understand the intuition behind this principle. thank you so much haha. I have a final for this class tomorrow morning and now I feel prepared for the langrange problem I'm betting is going to be on there. thanks again.

[cameronbaird]

I have used the Lagrange multiplier so much in the last few months and I finally understand why. Thank you Grant for yet another masterclass

[GaborGyebnar]

At 3:01 you say lambda equals y, so both of them need to be zero. But that's only true for x != 0, so "y=lambda" doesn't stand here. Actually, y=+-1, x=0, lambda=0 satisfy all three equations.

[alejrandom]

3:06 "since lambda equals y" Wait! that fact came from the assumption that x is not equal to zero, which means the first equation doesn't give you any information if x=0 and we have to work with only the last 2 equations. If we plug x=0 into the last one, we get y=+-1, now we plug x=0 and y=+-1 into the second equation and we get that lambda equals 0. (0, 1) and (0, -1) are still worth checking. I just saw the graph on the first video again and they happen to be local maxima and minima...

[garlic]

Thanks A lot!
One thing worth mentioning is that the 2 values you crossed out at 5:27 MINIMIZE the function on the constraint.

[gemacabero]

One question. When you analyze the case x = 0, why do you say that lambda has to equal y ? I didn't understand this part.

Lambda could equal zero and then y = 1. So you have x = 0 and y = 1, which do satisfy the third equation. Why would this be wrong? Thank you!

[Xilotl]

What a big help!! Now, I have a test to take on this in 1.5 hours....

[LeopoldoTejada]

I would say there is a mistake when checking for the possibility of x=0. It uses the fact that y=lambda, but this identity holds only if x is not zero. Actually the values x=0, y=1 and lambda=0 seem to be a valid solution, unless I made a mistake

[irtazaisawsome]

3Blue1Brown! I thought you sounded familiar. :)

[nitfanono]

3:32 IF X = 0 we said that it should satisfy the constraint why can't y be plus or minus 1

[kamitube]

Wish I found this video earlier. You just explained what my teacher took.a semester do so. Thank you

[mikeywatts]

It might be interesting to compare this solution with another method: equating dy/dx of the two given equations. (For f(x, y), we get d(f(x, y)) = 2xy*dx + x^2*dy. But since we're looking at a contour, d(f(x, y)) = 0; similarly, x^2+y^2 = 1 can be differentiated implicitly to get dy/dx = -x/y).This gives -x/y = -2xy/x^2. Solving, we get the same solution as in the video.

[hakeemnaa]

this is high school math, it is just a different notation
if you take the derivative of a function and equal it to zero, you will have no change nearby, so you are at max or min zone
here you don't equal it to zero, you equal it to another derivative, you put limda because the derivatives are not equal in magnitude
the problem will be that you will have many same directions of the derivative ( same slop or same ratio if change y over the change of x) between two functions even if x and y of one function are far away from the other function so you have to but the constraint

[pandicon]

So, assuming that the examined function has a maximum and a minimum, this method should always find them, right? Or are there some weird cases in which it would miss some solutions, making you miss for example a minimum?

[ethanlawrence]

Am I correct that following the Legrange Multiplier method only gives you values of x and y that make the function STATIONARY (so could be minimum or maximum).

[amaarquadri]

At 3:00, when you are analyzing the case where x = 0 you state that lambda = y.
But since x = 0, lambda = y no longer applies (that would be like saying 0 * lambda = 0 * y => lambda = y, which is false).
In fact, the x = 0, y = 1, lambda = 0 seems to be a solution to the system of equations.
It wouldn't be the maximum though, since the objective would evaluate to 0.

[hakeemnaa]

the direction is the slop, ratio of change y/ change of x
we equal the direction of both functions Y1/X1=Y2/X2
of course Y1≠ Y2 but we can say Y1= λY2, X1=λX2 ( change of x and y )


so

[piotrzalewski]

There is solution for x=0 (it is not true that y=lambda if x=0). The (two) solution(s) are [x, y, lambda] = [0, 1, 0] or [0, -1, 0] and both were easy to spot on the graph in the previous part. These are local (constraint) max. and min respectively. The reason (in the context) is that if grad(f)=0 (as it is for x=0) than grad(f)=0*grad(g) for every grad(g). The series is excellent anyway (as usual).

[alexdigg]

i have a similar problem except my g(x) is more like (x-7)^2 + (y-8)^2 = 18. When im taking the derivative of g(x) here do I need to subtract over the 18 or divide it over so that i have =1 like the example here?

[praneelmadhuvanesh]

huh? When checking the x=0 possibility, lambda would no longer equal y. So lambda could equal 0 and y could equal 1. Right?