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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 how to solve

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multivariable optimization
problems using the method of

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Lagrange multipliers, and I have
a nice problem here for

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you that can be solved
that way.

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So in this problem, we've got
an ellipse, the ellipse with

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equation x squared plus for
4y squared equals 4.

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So that's this ellipse, and we
want to inscribe a rectangle

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in it, so here I mean actually
a rectangle whose edges are

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parallel to the axes.

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So I want to inscribe a
rectangle in this ellipse, and

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among all such rectangles, I
want to find the one with the

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largest perimeter.

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So I want to find the maximal
perimeter of a rectangle that

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can be inscribed in
this ellipse.

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So why don't you have a go at
solving this problem, pause

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the video, work it out,
come back, and we

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can work it out together.

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So hopefully, you've had some
luck working on this problem.

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Let's get started on it.

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So one thing we need to start is
we need to figure out a way

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to sort of describe these
rectangles in a way that will

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let us describe their perimeter,
write down what

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their perimeter is.

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So a natural way to do that is
to call this upper right-hand

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corner of the rectangle, to
call it the point x, y.

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So x, y is going to be that
upper right-hand corner of the

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rectangle, and it's going to
be ranging over the region

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from this topmost point on the
ellipse down to this rightmost

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point on the ellipse, on this
quarter arc of the ellipse.

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So if that point is x, y, we
need to figure out what is the

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perimeter that we're
trying to optimize.

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So the perimeter here, P, which
is a function of x and

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y-- well, so x is this distance,
so the length of the

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horizontal edge of the rectangle
is 2x, and we've got

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two of those, so that's 4x from
the horizontal sides.

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And then this height is y, so
the length of the vertical

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side of the rectangle is 2y, so
the perimeter is going to

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be 4x plus 4y.

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So that's our objective function
that we're trying to

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optimize, that we're trying
to find the maximum of.

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And we also have the constraint
function g, which

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is x squared plus 4y squared,
and the constraint is that g

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is equal to 4.

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So we have the objective
function P--

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P of x, y-- and we have this
constraint function g, and so

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we want to write down some
equations using Lagrange

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multipliers whose solutions
will correspond to the

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possible maximum points of P.

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So what are those equations?

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Well, we need that the gradient
of P is parallel to

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the gradient g.

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So that means that we need Px
is equal to lambda times gx

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and Py is equal to lambda
times gy for

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some value of lambda.

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We need to find a value of
lambda that makes this true.

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And then also, our third
equation is the constraint

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equation, that g
is equal to 4.

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So what is Px equal lambda gx
translate to in our case?

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Let's just draw a line here.

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So in our case, Px is the x
partial derivative of 4x plus

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4y, so that's just 4, and gx, we
take the partial derivative

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with respect to x of x squared
plus 4y squared, and that's

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equal to 2x, so 4 is equal
to lambda times 2x.

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And from taking the y partial
derivatives, we have that the

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y partial derivative a P is 4,
Py is 4, and gy is going to be

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the y partial derivative of x
squared plus 4y squared, so

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that's 8y, so 4 equals lambda
times 8y, and we also have the

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constraint equation x squared
plus 4y squared equals 4.

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So we need to solve these three
equations, and we need

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to figure out which values of
x and y are the solutions.

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So I think the simplest way to
proceed here is to note that

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from the first equation and the
second equation, we can

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eliminate lambda between them,
and what we'll see is that x

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has to be exactly four times as
large as y for this to be

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true, for both of these
equations to be

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true at the same time.

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So we need x to be
equal to 4y.

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So from the first two equations,
we have that x is

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equal to 4y, and now we can
substitute that in to the

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constraint equation.

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So if x is 4y, then x squared
is 16y squared, so x squared

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plus 4y squared is
20y squared.

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So we have 20y squared
is equal to 4.

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And OK, so we can solve
this for y.

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We can divide by 20 and
take a square root,

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so we get that y--

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well, so y squared is equal to
1/5, so y is equal to plus or

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minus 1 over the square
root of 5.

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But remember, come back over
here, we've taken x, y to be

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the upper right-hand corner,
this first quadrant corner of

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our rectangle, so y is
always positive.

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So we had that y squared equals
1/5 and y is positive,

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so there's actually
only one root.

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We don't need to consider
the negative root.

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So over here, we know that
y is 1 divided by the

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square root of 5.

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OK, so that's y.

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Now what's x?

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Well, OK, so we solve for x in
terms of y, so x is equal to 4

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over the square root of 5.

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So Lagrange multipliers, when we
use the method of Lagrange

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multipliers, we get this one
possible point at which we

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have to check to
be the maximum.

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But remember that when you're
using Lagrange multipliers,

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you also have to worry about
the boundary of the region

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that you're interested in.

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So let's go look at
our picture again.

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So over on our picture, this
point x, y moved along the arc

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connecting the topmost point
of the ellipse to the

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rightmost point of
the ellipse.

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So we also have to look at the
perimeters when the point is

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the topmost point and when the
point is the rightmost point.

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Now, in those two cases, the
rectangle is a sort of

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degenerate rectangle, and when
x, y is this point 0, 1, it's

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sort of two copies of this
vertical line, this minor

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axis, and when x, y is the point
2, 0, then our rectangle

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just looks like the major
axis, which is

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that horizontal line.

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But we still have to check those
cases to see whether our

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function has a maximum
and what it is.

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So we need to compute the
objective function value at

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this point and we need to
compute it at those endpoints.

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So we need to look at P of--

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so this is our point 4 over the
square root of 5 comma 1

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over the square root of 5, and
we know that P of x, y is 4x

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plus 4y, so that's equal to 20
over the square root of 5,

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which we can also write as 4
times the square root of 5.

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And we also need to check those
two endpoints, so we

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need to check the point P of 0,
1, so that's 4, and we need

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to check the point P of
2, 0, so that's 8.

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So in order to find out what the
maximum value of P is, we

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need to compare the value of P
at the points given to us by

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Lagrange multipliers and at
the boundary points of the

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region, which in this case are
the endpoints of the arc.

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So we need to compare the
numbers 4 square root of 5, 4

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and 8, and indeed, 4
square root of 5 is

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the largest of these.

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So this is the largest, so
this is actually the

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maximum value, OK?

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So the maximum perimeter is
4 square root of 5 when

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rectangle has its upper
rightmost vertex at this

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point: 4 over square root of 5
comma 1 over square root of 5.

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So our rectangle's maximal
perimeter is 4 root 5, and

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that occurs when the upper
right-hand vertex is at the

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point 4 over root 5,
1 over root 5.

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So to quickly recap, we wanted
to apply the method of

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Lagrange multipliers
to this problem.

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So we chose to keep track of our
rectangles by their upper

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right-hand corner.

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And then that gave us-- the
perimeter was 4x plus 4y.

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That was our objective
function.

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And the constraint was that that
upper right-hand corner

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actually had to lie
on the ellipse.

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So then we set the gradients of
the two functions equal and

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solved the system of equations
that we get by having those--

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sorry, the gradients not to be
equal, but to be parallel.

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There's some constant multiple
lambda that appears.

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So we set the gradients to be
parallel to each other and the

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constraint equation to hold,
and we solved those three

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equation simultaneously
for x and y.

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And those equations gave us one
point that we had to check

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to be the maximum, and we also
needed to check points on the

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boundary of the region
in question.

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So here, those were just the
two points 0, 1 and 2, 0.

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So I'll end there.

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