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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

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to solve multivariable
optimization problems using

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the method of
Lagrange multipliers,

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and I have a nice
problem here for you

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that can be solved that way.

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

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the ellipse with equation
x squared plus 4 y squared

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equals 4.

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So that's this
ellipse, and we want

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to inscribe a rectangle
in it, so here I

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mean actually a rectangle whose
edges are parallel to the axes.

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So I want to inscribe a
rectangle in this ellipse,

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and among all such
rectangles, I want

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to find the one with
the largest perimeter.

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So I want to find
the maximal perimeter

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of a rectangle that can be
inscribed in this ellipse.

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

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pause the video, work
it out, come back,

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and we 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

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need to figure out a
way to sort of describe

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these rectangles in
a way that will let

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us describe their
perimeter, write down

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what their perimeter is.

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So a natural way to
do that is to call

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this upper right-hand corner
of the rectangle, the call it

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the point (x, y).

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So (x, y) is going to be
that upper right-hand corner

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of the rectangle, and
it's going to be ranging

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over the region from this
topmost point on the ellipse

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down to this rightmost
point on the ellipse,

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on this quarter
arc of the ellipse.

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So if that point
is (x, y), we need

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to figure out what is
the perimeter that we're

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trying to optimize.

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

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well, so x is this
distance, so the length

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of the horizontal edge
of the rectangle is 2x,

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and we've got two
of those, so that's

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4x from the horizontal sides.

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And then this height
is y, so the length

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of the vertical side
of the rectangle is 2y,

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so the perimeter is
going to be 4x plus 4y.

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

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that we're trying to
find the maximum of.

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And we also have the
constraint function

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g, which is x squared
plus 4 y squared,

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and the constraint is
that g is equal to 4.

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So we have the objective
function P-- P of x, y--

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and we have this
constraint function g,

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and so we want to write
down some equations using

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Lagrange multipliers whose
solutions will correspond

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to the 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

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is parallel to the gradient g.

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So that means that we need
P_x is equal to lambda times

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g_x and P_y is equal
to lambda times g_y

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for 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 does P_x equal lambda
g_x translate to in our case?

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Let's just draw a line here.

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

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

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with respect to x of x
squared plus 4y squared,

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and that's equal to 2x, so 4
is equal to lambda times 2x.

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And from taking the y
partial derivatives,

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we have that the y partial
derivative of P is 4, P_y is 4,

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and g_y is going to be the y
partial derivative of x squared

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plus 4 y squared, so that's 8y,
so 4 equals lambda times 8y,

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and we also have the
constraint equation x squared

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plus 4 y squared equals 4.

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So we need to solve
these three equations,

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and we need to figure out
which values of x and y

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are the solutions.

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So I think the simplest
way to proceed here

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is to note that from
the first equation

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and the second equation, we can
eliminate lambda between them,

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and what we'll see is that x
has to be exactly four times

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as large as y for
this to be true,

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for both of these equations
to be 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 equal to 4y,

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and now we can substitute that
in to the constraint equation.

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

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so x squared plus 4 y
squared is 20 y 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-- well, so
y squared is equal to 1/5,

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so y is equal to plus or minus
1 over the square root of 5.

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But remember, come back over
here, we've taken (x, y)

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to be the upper right-hand
corner, this first quadrant

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corner of 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

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by the 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,

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so x is equal to 4 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

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at which we 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)

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moved along the arc
connecting the topmost point

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of the ellipse to the
rightmost point of the ellipse.

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So we also have to
look at the perimeters

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when the point is
the topmost point

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and when the point is
the rightmost point.

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Now, in those two
cases, the rectangle

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is a sort of degenerate
rectangle, and when (x, y)

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is this point (0, 1),
it's sort of two copies

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of this vertical line, this
minor axis, and when (x, y)

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is the point (2, 0),
then our rectangle

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just looks like the major axis,
which is that horizontal line.

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But we still have
to check those cases

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to see whether our function
has a maximum and what it is.

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So we need to compute
the objective function

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value at this point and
we need to compute it

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at those endpoints.

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So we need to look at P
of-- so this is our point

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4 over the square
root of 5 comma

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

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is 4x plus 4y, so
that's equal to 20

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over the square root of 5,
which we can also write as 4

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times the square root of 5.

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And we also need to check
those two endpoints,

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

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and we need 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,

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we need to compare
the value of P

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at the points given to us
by Lagrange multipliers

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and at the boundary points of
the region, which in this case

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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 the largest of these.

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So this is the largest, so this
is actually the maximum value,

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OK?

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So the maximum perimeter
is 4 square root of 5

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when rectangle has its upper
rightmost vertex at this point:

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

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So our rectangle's
maximal perimeter

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is 4 root 5, and
that occurs when

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the upper right-hand vertex is
at the point 4 over root 5, 1

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over root 5.

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

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multipliers to this problem.

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So we chose to keep
track of our rectangles

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by their upper
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

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and solved the
system of equations

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that we get by having those--
sorry, the gradients not to be

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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

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to each other and the
constraint equation to hold,

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and we solved those three
equation simultaneously

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for x and y.

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And those equations
gave us one point

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that we had to check
to be the maximum,

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and we also needed
to check points

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on the 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.