Main content

## Multivariable calculus

### Course: Multivariable calculus > Unit 3

Lesson 5: Lagrange multipliers and constrained optimization- Constrained optimization introduction
- Lagrange multipliers, using tangency to solve constrained optimization
- Finishing the intro lagrange multiplier example
- Lagrange multiplier example, part 1
- Lagrange multiplier example, part 2
- The Lagrangian
- Meaning of the Lagrange multiplier
- Proof for the meaning of Lagrange multipliers

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Finishing the intro lagrange multiplier example

Working out the algebra for the final solution the the example from the previous two videos. Created by Grant Sanderson.

## Want to join the conversation?

- Is it true that lambda should never be 0 because it does not have any geometrical meaning?

(I get a little confused from3:12-3:27. If x = 0, in the first equation, we don't know for sure if y = lambda. Then in the second equation, either y =0, or lambda = 0, or both = 0. If only lambda = 0, in order to satisfy the third equation, y can be equal to =1 or -1.)(34 votes)- Just saw he added a pop-up note to point this out, but the problem is that (at least for me) it is not displayed when I'm watching it full-screen...(21 votes)

- There's a mistake in the video.
`y == lambda`

is the result of assumption that`x != 0`

. So when we consider`x == 0`

, we can't say that`y == lambda`

and hence the solution of`x^2 + y^2 = 0`

is impossible.

Instead we get this:

- Assume`x == 0`

- Then either`lambda == 0`

or`y == 0`

or both

- We know that`x^2 + y^2 == 0`

which gives us`y == +-1`

,`lambda == 0`

,`x == 0`

.

Now my question is can`lambda`

be actually equal to 0?(16 votes)- From the first video it looks like x=0, y=±1 are extreme points as well: [http://youtu.be/vwUV2IDLP8Q?t=100]. Just happened to be local extremums.(5 votes)

- Fun fact: Lagrange was actually Italian and not French(13 votes)
- Before I watched you work through it, I had taken the equation x^2+y^2=1, and solved for x^2 to get x^2=1-y^2, and then I took the definition for x^2, and substituted it into the multivariable function, to get a single variable function g(y), which equals y(1-y^2). At this point, I took the derivative of g(y), g'(y), and solved for when it equals zero. And I got the same solution for "y" that the other way produces. From there, I just solved the whole problem. Is there something about this method that conceals other details about the problem?(7 votes)
- Did the same thing :). My guess is that the constraints and the functions get even more multivariable and complicated and it's not always as easy or even possible to do so.(2 votes)

- Does this also work for finding the minimums and if so, how?(6 votes)
- Yes, you just have to find the points where f has a minimal value (as opposed to a maximal value).(3 votes)

- So what is the significance of the maximum value?(5 votes)
- Just comparing this to the previous unconstrained max / min problems where you get critical points using a similar approach, but use the Second partial derivative test to identify whether those critical points are max/mins.

Why is the constrained Lagrange max/min problems a quite different approach in "classifying" the critical points, not using anything like the second derivative test?

Thanks(5 votes) - Do the other 2 ordered pairs correspond to minimas ? (Under the given condition obviously)(4 votes)
- Hi I just wanted to point out that the edit for x = 0 is still wrong.

Indeed this would mean that lambda = 0. This however satisfies all three equations and means that grad f = 0. Indeed grad f = 0, along the entire line x = 0, which means the entire line consists of critical points. As two of these ((0,1) and (0,-1)) satisfy the constraint (and in fact are local min and max resp. on the constraint) indeed they need to be analysed along the other four Grant found.

Also don't forget to mention the case where grad g=0 as this in general does not satisfy the equation grad f=lambda*grad g. But indeed the zero-vector is parallel to any other vector. In this case however there are no such points that satisfy the constraint.

In this case Grants answer is correct, despite him missing to check these cases. However this could just as easily not have been the case.

A better idea for cases where (grad f, grad g) is a square matrix (as in this case) is to check when the determinant of the same matrix is zero. As this would include all cases mentioned above and of course the cases Grant found.(4 votes) - Can the lagrange multiplier take a negative value?(3 votes)
- Yes, it can take on any real value, including zero. The edit in the video is wrong.(2 votes)

## Video transcript

- [Instructor] So in the last two videos we were talking about this
constrained optimization problem where we want to maximize
a certain function on a certain set, the
set of all points x, y where x squared plus y squared equals one. And we ended up working
out through some nice, geometrical reasoning
that we need to solve this system of equations. So there's nothing left
to do but to just solve this system of equations. I will start with this
first one at the top, see what we can simplify. We notice there's an x term in each one so we'll go ahead and cancel those out, which is basically a way
of saying we're assuming that x is not zero, and we
can kind of return to that to see if x equals zero
could be as you wish. And so maybe we'll kind
of write that down. We're assuming x is not equal to zero in order to cancel out. And we kind of can revisit whether that could give us another
possible solution later, but that will be two times y
is equal to lambda times two, and from here, the twos can cancel out, no worries about two equaling zero, and we know that y equals lambda. So that's a nice, simplified
form for this equation. And for this next
equation, though we can use what we just found, that
y is equal to lambda to replace the lambda
that we see, and instead, if I replace this with
a y, what I'm gonna get is that x squared is equal
to y times two times y. So that's two times y squared. And I'll leave it in
that form because I see that in the next equation,
I see an x squared, I see a y squared, so it might be nice to be able to plug this guy right into it. So in that next equation, x squared, I'm gonna go ahead and
replace that with y squared, so that's two y squared plus y squared equals one. And then from there, simplifies
to three times y squared equals one, which in turn means y squared is equal to one third, and so y is equal to plus or minus the square root of one third. Great. So this gives us y, and
I'll go ahead and put a box around that, where we
have found what y must be. Now if y squared is equal to one third, then we look up here and say, mhm, two times y squared, that's
gonna be the same thing as two times one third,
so two times one third, so if x squared is equal to two thirds, what that implies is that x is equal to plus or minus the square root of two thirds. And then there we go, that's another, one of the solutions. And I could write down
what lambda is, right, I mean, in this case it's
easy, 'cause y equals lambda, but all we really want in
their final form are x and y, since that's gonna give us the answer to the original constraint problem. So this, this gives us what we want, and we just have that
pesky, little possibility that x equals zero to address, and for that we can take a look and say, if x equaled zero, you
know let's go through the possibility that
maybe that's one of the constrained solutions. Well, in this equation
that would make sense, since two times zero would equal zero. In this equation, that would
mean that we're setting zero equal to lambda times two times y. Well, since lambda
equals y, that would mean that for this side to equal zero, y would have to equal zero, so evidently, you know, if it was the
case that x equals zero, that would have to imply, from the second equation
that y equals zero, but if x and y both equal zero, this constraint can't be satisfied. So none of this is
possible, so we never even had to worry about this to start with, but it's something you do need to check, just every time you're
dividing by a variable, you're basically assuming
that it's not equal to zero. So, this right here gives
us four possible solutions, four possible values for x and y that satisfy this constraint,
and which potentially maximize this, and remember, when I say potentially maximize, the whole idea of this Lagrange Multiplier,
is that we were looking for where there's a point of tangency between the contour lines. So just to make it
explicit, the four points that we're dealing with, here
I'll right them all here. So, x could be the square
root of two thirds, square root of two thirds, and y could be the positive
square root of one third, and then we can basically just toggle, you know maybe x is the negative
square root of two thirds and y is still the positive
square root of one third, or maybe x is the positive square root of two thirds, and y is the negative
square root of one third, kind of monotonous, but just getting all of the different
possibilities on the table here. X is negative square root of two thirds, and then y is positive, no,
negative, that's the last one, square root of one third. So these are the four points
where the contour lines are tangent, and to
find which one of these maximizes our function, here let's go ahead and write
down our function again, it gets easy to forget. So the whole thing we're
doing is maximizing f of x, y, equals x squared times y. So let me just put that down again, we're looking at f of x, y is equal to x squared times y. So we could just plug these values in and see which one of them
is actually greatest. And the first thing to observe is, x squared is always gonna be positive, so if I plug in a negative value for y, right, if I plug in either
this guy here or this guy here, where the value for y is negative, the entire function would be negative. So I'm just gonna say
that neither of these can be the maximum because
it'll be some positive number, some x squared times a negative. Whereas I know that these
guys are gonna produce a positive number, and specifically, now if we plug in, if we
plug in f of let's say this top one, square root of two thirds, square root of one third, well, x squared is gonna be two thirds, and then y is square root of one third. And in fact that's gonna be the same as what we get plugging
in this other value. So either one of these
maximizes the function. It's got two different maximizing points and each one of them has a maximum value of two thirds times the
square root of one third. And that's the final answer. But I do want to emphasize,
that the takeaway here is not the specific
algebra that you work out going towards the end, but it's the whole, the whole idea of this
Lagrange Multiplier technique to find the gradient of one function, find the gradient of the
constraining function, and then set them
proportional to each other. That's the key takeaway, and
then the rest of it is just, you know making sure
that we check our work and go through the minute details, which is important, it has its place. And coming up, I'll go
through a few more examples.