JOEL LEWIS: Hi.
Welcome back to recitation.
In lecture, you've been
learning about using
the method of
Lagrange multipliers
to optimize functions of several
variables given a constraint.
So here's a problem that you
can practice this method on.
So I've got a function
f of x, y, z equals
x squared plus x plus 2 y
squared plus 3 z squared.
And what I'd like you to do is
find the maximum and minimum
values that this function
takes as the point (x, y, z)
moves around the unit sphere x
squared plus y squared plus z
squared equals 1.
So to optimize this function
given the constraint
x squared plus y squared
plus z squared equals 1.
So why don't you pause
the video, take some time
to work that out, come back,
and we can work it out together.
Hopefully you had some luck
working on this problem.
Let's have a go at it.
So remember that the
method of Lagrange
multipliers-- in order to
apply it-- what it says
is that when you have a
function being optimized
on some constraint
condition, what
you do to find the points where
the function could be maximum
or minimum is that first
you look for points where
the gradient of your
objective function
is parallel to the gradient
of your constraint function.
So what that means is you
take the partial derivatives
f_x, f_y, f_z, and you say f_x
has to be equal to lambda times
g_x, f_y has to be equal
to lambda times g_y,
and f_z has to be
equal to lambda times
g_z, for some lambda.
And then you solve
that system together
with the constraint equation.
And so the points
that are the solutions
of that system of
equations, those points
are your points that you have
to check for whether they're
the maximum or the minimum.
And also, sometimes you have
some boundary to your region
and you have to
check that as well.
So in this case, the sphere
doesn't have boundary.
Right?
So we don't have any
boundary conditions to check.
So we're going to have a
really straightforward problem
to solve where we
just have to look
at the partial derivatives.
So let's write down
that system of equations
that we have to solve.
So the partial derivative
of f with respect to x
is going to be 2x plus 1.
So we have to solve the
system 2x plus 1 equals--
and the partial derivative of
our constraint with respect
to x is 2x, so 2x plus 1 has
to equal lambda times 2x.
That's what we get from
the x-partial derivatives.
How about from the
y-partial derivatives?
The y-partial derivative
of f is going to be 4y.
So that has to be
equal to lambda
and the y-partial derivative
of the constraint equation
which is 2y.
And the z-partial
derivative of f is 6z.
So 6z has to be equal
to lambda times, well,
the z-partial derivative of
the constraint function, which
is 2z.
And we have the last equation
x squared plus y squared plus z
squared equals 1.
So we get four equations in
our variables x, y, and z,
plus this new parameter
lambda that we introduced.
And we want to
solve these to find
the points x, y, and z at
which these equations are all
satisfied.
And then, once we
get those points,
we have to test them
to see whether they
are the maximum or the
minimum or neither.
So OK.
So we have this
system of equations.
Now, this is a little
bit complicated.
It's not a system
of linear equations.
So we need to think about
ways that we can solve it.
And one thing that I
think we can do here,
is if you look at the
second and third equations,
you see that in the
second equation,
everything has a
factor of y in it.
So either y is equal to
0, or we can divide by it.
So from the second
equation, we have
that either y is
equal to 0, or we
can divide by y, in which case
we get lambda is equal to 2.
Similarly, from
the third equation,
we have that either z is equal
to 0, or we can divide by z
and we get lambda is equal to 3.
So from the third
equation, we have z
equals 0 or lambda equals 3.
So now we have a bunch
of possibilities, right?
So either we have y equals z
equals 0, or we have y equals 0
and lambda equals 3, or we have
lambda equals 2 and z equals 0.
Or, well, the other
possibility would be lambda
equals 2 and lambda equals
3, but that can't happen.
So we have three possibilities.
Three different ways that
this could be satisfied.
So let's go over
here and write down
what those possibilities are.
So case one, or maybe
I'll call it case a.
So the case a is when y is
equal to z is equal to 0.
So when y is equal to
z is equal to 0-- OK,
we need to find x still.
So let's look back
at our equations.
And when y is equal
to z is equal to 0,
well, we can solve our
constraint equation for x.
When y equals z equals 0, we
have that x squared equals 1.
So there are two possibilities.
The point (1, 0, 0), and
the point minus 1, 0, 0.
So this gives us,
in this case, we
have x equals 1 or
x equals minus 1.
So that gives us
the points (1, 0, 0)
and minus 1, 0, 0
that we're going
to have to check at the end.
All right.
So the second case is we
could have y equal to 0
and lambda equal to 3.
So in this case, let's go
back to our equations again.
So from lambda equals 3, we
have in our first equation
that 2x plus 1 equals 6x.
So 1 equals 4x or x equals 1/4.
So this implies over
here that x equals 1/4.
And now, we still
need to find z.
So if we go back to our
constraint equation here,
we have that x is a
quarter and y is 0.
So that means 1/16 plus
z squared equals 1.
So z has to be the square
root of 15/16, plus or minus.
And z is equal to
plus or minus--
so that we can also write
that as the square root
of 15 over 4.
So this also gives us
two points to check.
The points are 1/4, 0, the
square root of 15 over 4.
And 1/4, 0, minus square
root of 15 over 4.
And finally, we
have our third case.
So our third case is
when lambda is equal to 2
and z is equal to 0.
So again, let's go back
over to our equation.
So when lambda equals 2
in the first equation,
we have 2x plus 1 equals 4x.
So 2x equals 1 or x is 1/2.
So this gives us
x equals a half.
And now if you take z
equals 0 and x equals 1/2,
we can take that down to
our constraint equation.
And we get a quarter
plus y squared equals 1,
so y is a square root of 3/4.
So y equals plus or minus
square root of 3 over 2.
And this gives us two points.
1/2, square root of 3 over 2, 0.
And 1/2, minus square
root of 3 over 2, 0.
Those were our three cases.
We've solved each of them.
We've solved each of
them all the way down
to finding the points
that they lead to.
Now remember, we said
already that there's
no boundary to this region.
It's just the sphere.
It has no edges.
So these are the only
points we have to check.
We have to check
these six points.
What do we have
to check them for?
Well, we have to look
at the value of f
at each of these six points.
And we want to figure
out where f is maximized
and where f is minimized,
and these six points are
the only points where
that could happen,
where f could be
maximized or minimized.
So we just have to evaluate
our objective function
f at these six points
and find the largest
value and the smallest value.
So let's do that.
So our objective
function, remember, it's
all the way back over here.
It's this function x
squared plus x plus 2 y
squared plus 3 z squared.
OK.
So let's look at the value of
that function at these point.
So x squared plus
x plus 2 y squared
plus 3 z squared at the point
1, 0, that's just equal to 2.
So I'm going to write the
function values just off
to the side of the points here.
So this gives me the value 2.
And I'm going to circle them.
So the point (1, 0, 0)
gives me the value 2.
The point minus 1, 0,
0-- so that's x squared
is 1, plus x is minus
1-- so that's 1 minus 1
is 0-- and then the y
and z terms are both 0.
So at the point minus 1, 0,
0, the function value is 0.
I'm going to circle that.
Oh boy.
OK, so at these points-- at
the point 1/4, 0, square root
of 15 over 4, and 1/4, 0, minus
square root of 15 over 4--
I'm going to cheat and look
at what I wrote down already.
So you could do the
arithmetic yourself,
but I think it's not
that hard to work out
that in both of these cases, the
function value that you get out
is 25 over 8.
I'm not going to do the
arithmetic right now.
But you can double-check
that for yourself.
And at these last two
points-- the points 1/2,
root 3 over 2, 0, and
1/2, minus root 3 over 2,
0-- the function has the same
value at both of those points.
That value is 9/4.
Yeah, so 25 over 8 was the
value at both of these points,
and 9/4 is the value of
both of these points.
So now, to find the maximum
value of the function
and the minimum value
of the function,
we just look at the values
that we got and say,
which of these is biggest and
which of these is smallest?
And in our case, it's easy
to see that 0 is the minimum.
You know, all the other
values are positive,
so 0 is the minimum.
So our minimum value of f is
0 at the point minus 1, 0, 0.
And if you just compare the
values 2 and 25/8 and 9/4,
25/8 is the largest.
This is bigger than 3, whereas
both of those are less than 3,
for example.
This is one easy
way to see that.
So the max of f is
25/8, and that's
achieved at the points
1/4, 0, plus or minus
square root of 15 over 4.
So there you have it.
The method of
Lagrange multipliers.
We just followed exactly
the strategy that we have.
So you start out and you
have an objective function
and a constraint function.
And so what do you do?
You write down their
partial derivatives
and you come up with
this system of equations.
So this system of equations
that you get by setting,
you know, f_x equal to lambda
g_x, f_y equal to lambda g_y,
f_z equals lambda g_z, and
your constraint equation g
equals some constant.
So then the one part of this
procedure that isn't just
a recipe is that you need to
solve this system of equations,
but sometimes that can be hard.
So in this case, there were
a couple of observations
that we could make from the
second and third equations that
made it relatively
straightforward to do.
And that gave us some cases.
And then in each
of those cases, we
were able to completely solve
for the points x, y, and z.
Now we also could solve for the
associated values of lambda,
but lambda isn't
important to us.
It doesn't affect f.
We can forget about it
as soon as we found it,
once we found x, y, and z.
So we were able to solve.
In this case, we got
six points of interest.
And then you just
look at the value
of your objective
function at those points.
So that was what I wrote
down in these circles.
So you look at the value
of the objective function.
And to find the maximum
value of the function,
you just look at which
of those is largest.
Now sometimes-- not in this
problem, but in other problems,
you'll also have to check--
if the region has a boundary,
you'll also have to check for
possible maxima and minima
on the boundary of the region.
But a sphere doesn't
have any edges,
so it doesn't have any boundary.
So we don't have to
worry about that.
So that's how we apply the
method of Lagrange multipliers
to this problem.
And how you can apply it
to other problems as well.
I'll end there.
