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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've been learning
about using the

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method of Lagrange multipliers
to optimize functions of

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several variables given
a constraint.

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So here's a problem that you can
practice this method on.

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So I've got a function f of x,
y, z equals x squared plus x

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plus 2y squared plus
3z squared.

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And what I'd like you to do is
find the maximum and minimum

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values that this function takes
as the point x, y, z

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moves around the unit sphere x
squared plus y squared plus z

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squared equals 1.

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So to optimize this function
given the constraint x squared

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plus y squared plus z
squared equals 1.

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So why don't you pause the
video, take some time to work

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that out, come back, and we
can work it out together.

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Hopefully you had some luck
working on this problem.

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Let's have a go at it.

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So remember that the method
of Lagrange multipliers--

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in order to apply it-- what it
says is that when you have a

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function being optimized on
some constraint condition,

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what you do to find the points
where the function could be

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maximum or minimum is that
first you look for points

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where the gradient of your
objective function is parallel

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to the gradient of your
constraint function.

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So what that means is you take
the partial derivatives fx,

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fy, fz, and you say fx has to
be equal to lambda times gx,

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fy has to be equal to lambda
times gy, and fz has to be

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equal to lambda times
gz, for some lambda.

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And then you solve that
system together with

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the constraint equation.

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And so the points that are the
solutions of that system of

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equations, those points are your
points that you have to

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check for whether they're the
maximum or the minimum.

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And also, sometimes you have
some boundary to your region

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and you have to check
that as well.

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So in this case, the sphere
doesn't have boundary.

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Right?

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So we don't have any boundary
conditions to check.

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So we're going to have a really
straightforward problem

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to solve where we just have
to look at the partial

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derivatives.

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So let's write down that system
of equations that we

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have to solve.

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So the partial derivative of f
with respect to x is going to

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be 2x plus 1.

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So we have to solve the system
2x plus 1 equals.

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And the partial derivative of
our constraint with respect to

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x is 2x, so 2x plus 1 has to
equal lambda times 2x.

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That's what we get from the
x-partial derivatives.

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How about from the y-partial
derivatives?

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The y-partial derivative
of f is going to be 4y.

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So that has to be equal to
lambda and the y-partial

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derivative of the constraint
equation which is 2y.

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And the z-partial derivative
of f is 6z.

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So 6z has to be equal to
lambda times, well the

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z-partial derivative
of the constraint

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function which is 2z.

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And we have the last equation x
squared plus y squared plus

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z squared equals 1.

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So we get four equations in
our variables x, y, and z,

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plus this new parameter lambda
that we introduced.

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And we want to solve these to
find the points x, y, and z at

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which these equations
are all satisfied.

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And then, once we get those
points, we have to test them

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to see whether they
are the maximum or

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the minimum or neither.

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So OK.

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So we have this system
of equations.

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Now, this is a little
bit complicated.

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It's not a system of
linear equations.

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So we need to think about ways
that we can solve it.

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And one thing that I think we
can do here, is if you look at

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the second and third equations,
you see that in the

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second equation, everything
has a factor of y in it.

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So either y is equal to 0,
or we can divide by it.

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So from the second equation, we
have that either y is equal

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to 0, or we can divide by y, in
which case we get lambda is

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equal to 2.

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Similarly, from the third
equation, we have that either

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z is equal to 0, or we can
divide by z and we get lambda

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is equal to 3.

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So from the third equation
we have z equals 0 or

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lambda equals 3.

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So now we have a bunch of
possibilities, right?

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So either we have y equals z
equals 0, or we have y equals

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0 and lambda equals 3, or
we have lambda equals 2

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and z equals 0.

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Or well, the other possibility
would be lambda equals 2 and

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lambda equals 3, but
that can't happen.

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So we have three
possibilities.

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Three different ways that
this could be satisfied.

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So let's go over here and
write down what those

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possibilities are.

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So case one, or maybe
I'll call it case a.

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So the case a is when y is
equal to z is equal to 0.

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So when y is equal to
z is equal to 0--

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OK, we need to find x's still.

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So let's look back
at our equations.

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And when y is equal to z is
equal to 0, well we can solve

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our constraint equation for x.

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When y equals z equals 0, we
have that x squared equals 1.

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So there are two
possibilities.

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The point 1, 0, 0, and the
point minus 1, 0, 0.

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So this gives us, in this case,
we have x equals 1 or x

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equals minus 1.

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So that gives us the points 1,
0, 0, and minus 1, 0, 0 that

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we're going to have to
check at the end.

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All right.

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So the second case is we could
have y equal to 0 and lambda

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equal to 3.

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So in this case, let's go back
to our equations again.

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So from lambda equals 3, we
have in our first equation

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that 2x plus 1 equals 6x.

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So 1 equals 4x or
x equals 1/4.

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So this implies over here
that x equals 1/4.

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And now, we still
need to find z.

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So if we go back to our
constraint equation here, we

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have that x is a quarter
and y is 0.

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So that means 1/16 plus
z squared equals 1.

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So z has to be the square root
of 15/16, plus or minus.

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And z is equal to plus or minus,
so that we can also

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write that as the square
root of 15 over 4.

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So this also gives us
two points to check.

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The points are 1/4, 0, the
square root of 15 over 4.

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And 1/4, 0, minus square
root of 15 over 4.

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And finally, we have
our third case.

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So our third case is when lambda
is equal to 2 and z is

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equal to 0.

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So again, let's go back
over to our equation.

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So when lambda equals 2 in the
first equation, we have 2x

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plus 1 equals 4x.

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So 2x equals 1 or x is 1/2.

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So this gives us x
equals a half.

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And now if you take z equals 0
and x equals 1/2, we can take

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that down to our constraint
equation.

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And we get a quarter plus y
squared equals 1, so y is a

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square root of 3/4.

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So y equals plus or minus
square root of 3 over 2.

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And this gives us two points.

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1/2, square root
of 3 over 2, 0.

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And 1/2, minus square
root of 3 over 2, 0.

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Those were our three cases.

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We've solved each of them.

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We've solved each of them all
the way down to finding the

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points that they lead to.

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Now remember, we said already
that there's no boundary to

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this region.

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It's just the sphere.

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It has no edges.

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So these are the only points
we have to check.

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We have to check these
six points.

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What do we have to
check them for?

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Well, we have to look at the
value of f at each of these

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six points.

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And we want to figure out where
f is maximized and where

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f is minimized, and these six
points are the only points

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where that could happen,
where f could be

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maximized or minimized.

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So we just have to evaluate our
objective function f at

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these six points and find
the largest value and

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the smallest value.

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So let's do that.

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So our objective function,
remember, it's all the way

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back over here.

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It's this function x squared
plus x plus 2y

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squared plus 3z squared.

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OK.

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So let's look at the value of
that function at these point.

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So x squared plus x plus 2y
squared plus 3z squared at the

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point 1, 0, that's
just equal to 2.

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So I'm going to write the
function values just off to

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the side of the points here.

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So this gives me the value 2.

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And I'm going to circle them.

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So the point 1, 0, 0 gives
me the value 2.

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The point minus 1, 0, 0--

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so that's x squared is one, plus
x is minus 1, so that's 1

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minus 1 is 0, and then the
y and z terms are both 0.

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So at the point minus 1, 0, 0,
the function value is 0.

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I'm going to circle that.

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Oh boy.

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OK, so at these points--

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at the point 1/4, 0, square root
of 15 over 4, and 1/4, 0,

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minus square root
of 15 over 4--

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I'm going to cheat and look at
what I wrote down already.

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So you could do the arithmetic
yourself, but I think it's not

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that hard to work out that in
both of these cases, the

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function value that you
get out is 25 over 8.

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I'm not going to do the
arithmetic right now.

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But you can double-check
that for yourself.

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And at these last two points--
the points 1/2, root 3 over 2,

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0, and 1/2, minus root
3 over 2, 0--

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the function has the same value
at both of those points.

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That value is 9/4.

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Yeah, so 25 over 8 was the value
at both of these points,

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and 9/4 is the value of
both of these points.

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So now, to find the maximum
value of the function and the

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minimum value of the function,
we just look at the values

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that we got and say, which of
these is biggest and which of

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these is smallest?

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And in our case, it's easy to
see that 0 is the minimum.

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You know, all the
other values are

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positive, so 0 is the minimum.

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So our minimum value of f is 0
at the point minus 1, 0, 0.

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And if you just compare the
values 2 and 25/8 and 9/4,

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25/8 is the largest. This is
bigger than 3, whereas both of

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those are less than
3, for example.

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This is one easy way
to see that.

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So the max of f is 25/8, and
that's achieved at the points

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1/4, 0, plus or minus square
root of 15 over 4.

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So there you have it.

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The method of Lagrange
multipliers.

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We just followed exactly the
strategy that we have.

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So you start out and you have
an objective function and a

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constraint function.

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And so what do you do?

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You write down their partial
derivatives and you come up

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with this system of equations.

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So this system of equations that
you get by setting, you

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know, fx equal to lambda gx,
fy equal to lambda gy, fz

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equals lambda gz, and your
constraint equation g equals

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some constant.

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So then the one part of this
procedure that isn't just a

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recipe is that you need to
solve this system of

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equations, but sometimes
that can be hard.

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So in this case, there were a
couple of observations that we

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could make from the second and
third equations that made it

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relatively straightforward
to do.

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And that gave us some cases.

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And then in each of those
cases, we were able to

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completely solve for the
points x, y, and z.

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Now we also could solve for
the associated values of

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lambda, but lambda isn't
important to us.

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It doesn't affect f.

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00:12:31,690 --> 00:12:34,990
We can forget about it as soon
as we found it, once we found

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x, y, and z.

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00:12:36,320 --> 00:12:37,490
So we were able to solve.

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00:12:37,490 --> 00:12:40,330
In this case, we got six points
of interest. And then

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00:12:40,330 --> 00:12:42,690
you just look at the value of
your objective function at

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00:12:42,690 --> 00:12:43,470
those points.

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00:12:43,470 --> 00:12:46,280
So that was what I wrote
down in these circles.

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00:12:46,280 --> 00:12:49,440
So you look at the value of
the objective function.

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00:12:49,440 --> 00:12:51,810
And to find the maximum value of
the function, you just look

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00:12:51,810 --> 00:12:53,300
at which of those is largest.

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00:12:53,300 --> 00:12:54,220
Now sometimes--

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00:12:54,220 --> 00:12:55,751
not in this problem, but in
other problems, you'll also

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00:12:55,751 --> 00:12:57,640
have to check--

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00:12:57,640 --> 00:13:00,420
if the region has a boundary,
you'll also have to check for

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00:13:00,420 --> 00:13:03,180
possible maxima and minima on
the boundary of the region.

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00:13:03,180 --> 00:13:05,160
But a sphere doesn't have
any edges, so it

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00:13:05,160 --> 00:13:06,070
doesn't have any boundary.

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00:13:06,070 --> 00:13:08,020
So we don't have to
worry about that.

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00:13:08,020 --> 00:13:10,770
So that's how we apply the
method of Lagrange multipliers

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00:13:10,770 --> 00:13:11,690
to this problem.

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00:13:11,690 --> 00:13:14,370
And how you can apply it to
other problems as well.

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00:13:14,370 --> 00:13:16,080
I'll end there.

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00:13:16,080 --> 00:13:16,263