WEBVTT
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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.