JOEL LEWIS: Hi.
Welcome back to recitation.
In lecture, you've
been learning about how
to solve multivariable
optimization problems using
the method of
Lagrange multipliers,
and I have a nice
problem here for you
that can be solved that way.
So in this problem,
we've got an ellipse,
the ellipse with equation
x squared plus 4 y squared
equals 4.
So that's this
ellipse, and we want
to inscribe a rectangle
in it, so here I
mean actually a rectangle whose
edges are parallel to the axes.
So I want to inscribe a
rectangle in this ellipse,
and among all such
rectangles, I want
to find the one with
the largest perimeter.
So I want to find
the maximal perimeter
of a rectangle that can be
inscribed in this ellipse.
So why don't you have a go
at solving this problem,
pause the video, work
it out, come back,
and we can work it out together.
So hopefully, you've had some
luck working on this problem.
Let's get started on it.
So one thing we
need to start is we
need to figure out a
way to sort of describe
these rectangles in
a way that will let
us describe their
perimeter, write down
what their perimeter is.
So a natural way to
do that is to call
this upper right-hand corner
of the rectangle, the call it
the point (x, y).
So (x, y) is going to be
that upper right-hand corner
of the rectangle, and
it's going to be ranging
over the region from this
topmost point on the ellipse
down to this rightmost
point on the ellipse,
on this quarter
arc of the ellipse.
So if that point
is (x, y), we need
to figure out what is
the perimeter that we're
trying to optimize.
So the perimeter here, P, which
is a function of x and y--
well, so x is this
distance, so the length
of the horizontal edge
of the rectangle is 2x,
and we've got two
of those, so that's
4x from the horizontal sides.
And then this height
is y, so the length
of the vertical side
of the rectangle is 2y,
so the perimeter is
going to be 4x plus 4y.
So that's our objective function
that we're trying to optimize,
that we're trying to
find the maximum of.
And we also have the
constraint function
g, which is x squared
plus 4 y squared,
and the constraint is
that g is equal to 4.
So we have the objective
function P-- P of x, y--
and we have this
constraint function g,
and so we want to write
down some equations using
Lagrange multipliers whose
solutions will correspond
to the possible
maximum points of P.
So what are those equations?
Well, we need that
the gradient of P
is parallel to the gradient g.
So that means that we need
P_x is equal to lambda times
g_x and P_y is equal
to lambda times g_y
for some value of lambda.
We need to find a value of
lambda that makes this true.
And then also, our third
equation is the constraint
equation, that g is equal to 4.
So what does P_x equal lambda
g_x translate to in our case?
Let's just draw a line here.
So in our case, P_x is the x
partial derivative of 4x plus
4y, so that's just 4, and g_x,
we take the partial derivative
with respect to x of x
squared plus 4y squared,
and that's equal to 2x, so 4
is equal to lambda times 2x.
And from taking the y
partial derivatives,
we have that the y partial
derivative of P is 4, P_y is 4,
and g_y is going to be the y
partial derivative of x squared
plus 4 y squared, so that's 8y,
so 4 equals lambda times 8y,
and we also have the
constraint equation x squared
plus 4 y squared equals 4.
So we need to solve
these three equations,
and we need to figure out
which values of x and y
are the solutions.
So I think the simplest
way to proceed here
is to note that from
the first equation
and the second equation, we can
eliminate lambda between them,
and what we'll see is that x
has to be exactly four times
as large as y for
this to be true,
for both of these equations
to be true at the same time.
So we need x to be equal to 4y.
So from the first two equations,
we have that x is equal to 4y,
and now we can substitute that
in to the constraint equation.
So if x is 4y, then x
squared is 16 y squared,
so x squared plus 4 y
squared is 20 y squared.
So we have 20y
squared is equal to 4.
And OK, so we can
solve this for y.
We can divide by 20
and take a square root,
so we get that y-- well, so
y squared is equal to 1/5,
so y is equal to plus or minus
1 over the square root of 5.
But remember, come back over
here, we've taken (x, y)
to be the upper right-hand
corner, this first quadrant
corner of our rectangle,
so y is always positive.
So we had that y squared
equals 1/5 and y is positive,
so there's actually
only one root.
We don't need to consider
the negative root.
So over here, we know
that y is 1 divided
by the square root of 5.
OK, so that's y.
Now what's x?
Well, OK, so we solve
for x in terms of y,
so x is equal to 4 over
the square root of 5.
So Lagrange multipliers, when
we use the method of Lagrange
multipliers, we get
this one possible point
at which we have to
check to be the maximum.
But remember that when you're
using Lagrange multipliers,
you also have to worry about
the boundary of the region
that you're interested in.
So let's go look at
our picture again.
So over on our picture,
this point (x, y)
moved along the arc
connecting the topmost point
of the ellipse to the
rightmost point of the ellipse.
So we also have to
look at the perimeters
when the point is
the topmost point
and when the point is
the rightmost point.
Now, in those two
cases, the rectangle
is a sort of degenerate
rectangle, and when (x, y)
is this point (0, 1),
it's sort of two copies
of this vertical line, this
minor axis, and when (x, y)
is the point (2, 0),
then our rectangle
just looks like the major axis,
which is that horizontal line.
But we still have
to check those cases
to see whether our function
has a maximum and what it is.
So we need to compute
the objective function
value at this point and
we need to compute it
at those endpoints.
So we need to look at P
of-- so this is our point
4 over the square
root of 5 comma
1 over the square root of 5,
and we know that P of x, y
is 4x plus 4y, so
that's equal to 20
over the square root of 5,
which we can also write as 4
times the square root of 5.
And we also need to check
those two endpoints,
so we need to check the
point P of 0, 1, so that's 4,
and we need to check the
point P of 2, 0, so that's 8.
So in order to find out what
the maximum value of P is,
we need to compare
the value of P
at the points given to us
by Lagrange multipliers
and at the boundary points of
the region, which in this case
are the endpoints of the arc.
So we need to compare the
numbers 4 square root of 5, 4
and 8, and indeed, 4 square root
of 5 is the largest of these.
So this is the largest, so this
is actually the maximum value,
OK?
So the maximum perimeter
is 4 square root of 5
when rectangle has its upper
rightmost vertex at this point:
4 over square root of 5 comma
1 over square root of 5.
So our rectangle's
maximal perimeter
is 4 root 5, and
that occurs when
the upper right-hand vertex is
at the point 4 over root 5, 1
over root 5.
So to quickly recap, we wanted
to apply the method of Lagrange
multipliers to this problem.
So we chose to keep
track of our rectangles
by their upper
right-hand corner.
And then that gave us-- the
perimeter was 4x plus 4y.
That was our objective function.
And the constraint was that
that upper right-hand corner
actually had to
lie on the ellipse.
So then we set the gradients
of the two functions equal
and solved the
system of equations
that we get by having those--
sorry, the gradients not to be
equal, but to be parallel.
There's some constant
multiple lambda that appears.
So we set the gradients
to be parallel
to each other and the
constraint equation to hold,
and we solved those three
equation simultaneously
for x and y.
And those equations
gave us one point
that we had to check
to be the maximum,
and we also needed
to check points
on the boundary of the
region in question.
So here, those were just the
two points (0, 1) and (2, 0).
So I'll end there.
