-
PROFESSOR: --here.
-
I have excuse two
people for being sick.
-
But I haven't
excused anybody else.
-
You are not the complete group.
-
I would like to take
attendance as soon as possible.
-
Would you mind starting
an attendance sheet?
-
STUDENT: We already got one.
-
STUDENT: We already got one.
-
STUDENT: Yeah, we
already got one.
-
PROFESSOR: Oh, you
already have one.
-
OK.
-
-
I understand having
to struggle with snow,
-
but you are expected
to come here.
-
And I don't want to punish
the people who don't make it.
-
I want to reward the people
who make it every time.
-
That's the principle behind
perfect attendance for this.
-
All right.
-
Today we are going to
cover something new.
-
It is new and it's not new.
-
It's an extension of
the ideas in 11.7,
-
which were finding extrema
of functions of the type
-
z equals f of xy, and
classifying those.
-
-
In 11.8, we provide a
very specific method.
-
That's Lagrange multipliers.
-
-
Of finding extrema,
you struggled.
-
Well, you didn't
struggle, but it
-
wasn't easy to find those
absolute extrema at every time.
-
The Lagrange multipliers
are going to help you.
-
So practically, what
should we assume
-
to know that the function of
two variables that we deal with
-
is c1.
-
Sometimes I assume it's smooth.
-
What do we need?
-
We need the derivatives.
-
Derivatives.
-
Derivative exist
and are continuous.
-
-
I assume differentiability.
-
-
OK?
-
And what else do I assume?
-
I assume that you have a
constraint that is also smooth.
-
-
Let's say g of xy equals c.
-
-
Do you remember?
-
We talked about constraints
last time on Tuesday as well.
-
So practically,
x and y are bound
-
to be together by some sort of
agreement, contract, marriage.
-
They depend on one another.
-
They cannot leave
this constraint.
-
And last time, I
really don't remember
-
what problem I took last time.
-
But we had something like, given
the function f of xy-- that
-
was nice and smooth--
find the absolute maximum
-
and the absolute
minimum of that function
-
inside the-- or on
the closed disk.
-
Remember that?
-
We had the closed
disk, x squared
-
plus y squared less
than or equal to 1.
-
And we said, let's
find-- somebody gives you
-
this very nice function.
-
We found the critical
point inside.
-
And we said, that's it.
-
Relative max or relative min.
-
Maybe we have more
depending on the function.
-
What was crucial for us to
do-- to study the extrema that
-
could come from the boundary.
-
And in order for them to
come from the boundary,
-
we played this little game.
-
We took the boundary.
-
We said, that's the circle.
-
X squared plus y
squared equals 1.
-
We pulled out the
y in terms of x
-
and brought it back in
the original expression,
-
z equals f of xy.
-
Since y would depend on x as y
squared is 1 minus x squared,
-
we plugged in and we got
a function of x only.
-
For that function of x only
on the boundary, we said,
-
we look for those critical
points for the function.
-
It was a little bit of
time-consuming stuff.
-
Critical points.
-
That would give you
relative max or min
-
for that function
on the boundary.
-
Plus, we said, but that function
has n points in the domain,
-
because the domain
would be for x
-
between minus 1 and 1--
inclusively minus 1 and 1.
-
So those minus 1 and 1's
as endpoints can also
-
generate absolute max and min.
-
So we made a table of
all the possible values,
-
including all the critical
points and the values
-
at the endpoints.
-
We said, whoever's the
tallest guy over here's
-
gonna be the maximum.
-
Whoever's the smallest
one will be the minimum.
-
And that was what the
philosophy was before.
-
Now we have to find a
different method, which
-
is providing the same solution,
but it's more systematic
-
in approach and is
based on a result that
-
was due to Lagrange, one
of the-- well, the fathers.
-
The fathers were
Euler and Leibniz.
-
Lagrange had lots
of contributions
-
to physics, mechanics
especially, and calculus.
-
So he's also a father.
-
As a father, he came up
with this beautiful theorem,
-
that says, if you have
these conditions satisfied
-
and if has-- if f has
an extrema already--
-
we know that has an extrema.
-
-
At some point, P0 of x0 y0 along
the curve-- the boundary curve.
-
Let's call this boundary curve
as script C. Do you understand?
-
This is not an l.
-
I don't know how to denote.
-
Script C. Script C. How
do you draw a script C?
-
Let's draw it like that.
-
I'm not an l.
-
OK?
-
C from curve.
-
Then there exists-- I
taught you the sign.
-
There exists a lambda--
real number-- such
-
that the gradients are parallel.
-
What?
-
The gradient of f
of x0 y0 would be
-
parallel of a proportionality
factor, lambda,
-
to the gradient of g,
the constraint function.
-
So we have two
Musketeers here that
-
matter-- the gradient of
the original function,
-
the one you want to
optimize, and the gradient
-
of the constraint function
as a function of x
-
and y at the point.
-
And we claim that
at that point, we
-
have an extremum of some sort.
-
Then something magical happens.
-
There is a lambda-- there
is a proportionality
-
between those two gradients.
-
So you say something magical
happens-- that the gradients
-
will be in the same direction.
-
And the proportionality factor
is this beautiful lambda.
-
-
If Mr. g-- this is a tricky
thing you have to make sure
-
happens.
-
If Mr. g, let's say, is-- at
its 0y0 is different from 0.
-
Because if it is equal to 0,
well, then it's gonna be crazy.
-
We will have 0 equals 0 for
any lambda multiplication here.
-
So that would complicate
things, and you would
-
get something that's lost.
-
So how do I view the procedure?
-
How do I get the lambda?
-
Once I grab this lambda,
I think I would be done.
-
Because once I
grab the lambda, I
-
could figure out who the x0,
y0 are from the equations.
-
So I have this feeling I need a
procedure, I need an algorithm.
-
Engineers are more
algorithmically oriented
-
than us mathematicians.
-
And this is what I appreciate
mostly about engineers.
-
They have a very
organized, systematic mind.
-
So if I were to
write an algorithm,
-
a procedure for the
method, I would say,
-
assume that f and
g are nice to you.
-
You don't say that.
-
Don't write that.
-
Now assume that f and g
satisfy Lagrange's theorem.
-
Satisfy the conditions
of Lagrange's theorem.
-
-
OK?
-
The notion, by the way, has
nothing to do with Calc 3.
-
But the notion of
Lagrangian and Hamiltonian
-
are something you are
learning in engineering.
-
And the Lagrangian is a
product of Mr. Lagrange.
-
So he's done a lot for
science in general, not just
-
for calculus, for mathematics,
for pure mathematics.
-
OK.
-
-
In that case, what
do you need to do?
-
Step one.
-
-
You need to recover that.
-
So solve for x0, y0, and
lambda the following system.
-
What does it mean the two
gradients are parallel
-
to each other?
-
This fella over here is
going to be what vector?
-
f of xy-- f sub y.
-
This gal over here will
be g sub x, g sub y.
-
For them to be proportional,
you should have this,
-
then-- f sub x equals g
sub x times the lambda.
-
Right?
-
And f sub y equals g
sub y times the lambda
-
for the same lambda, your hero.
-
So both coordinates have to be
multiplied by the same lambda
-
to get you the other
partial velocities.
-
STUDENT: And it has to be
evaluated at that point?
-
PROFESSOR: Yes.
-
So you're gonna solve for--
you're gonna solve this system,
-
and you are going to
get-- and I'm sorry.
-
With a constraint.
-
And with the absolute constraint
that you have at g of xy
-
equals c, because
these guys are married.
-
They always are in
this relationship.
-
And from all the
information of the system,
-
you're going to get a-- not one,
maybe several values of lambda,
-
you get values of lambda, x0,
y0, that satisfy the system.
-
You have to satisfy all
the three constraints, all
-
the three equations.
-
I'm going to put them
in bullets, red bullets.
-
You don't have colors, but I do.
-
So.
-
And then at all these points
that we found in step one,
-
step two.
-
Step two I'm going
to erase here.
-
For all the points-- x0, y0,
and lambda 0-- you got step one.
-
You have to evaluate
the f function.
-
Evaluate f at those x0, y0,
lambda 0 we got 4 lambda 0.
-
And get to compare values
in a table just like before.
-
-
See all the points.
-
All points will
give you an idea who
-
is going to be the
absolute max, absolute min.
-
-
And I'm just going to go ahead
and solve one typical example
-
for your better understanding.
-
You know, it's not solved
in the book by both methods.
-
But I'm thinking
since I'm teaching you
-
how to apply the Lagrange
theorem today and do the step
-
one, step two
procedure for Lagrange
-
multipliers I'm going to solve
it with Lagrange multipliers
-
first.
-
And the same problem, I'm
going to solve it in the spirit
-
that we have employed
last time in 11.7.
-
And then I'm going
to ask you to vote
-
which method is easier for you.
-
And I'm really curious,
because of course, I
-
can predict what theorems
I'm going to cover.
-
And I can predict
the results I'm
-
going to get in the exercises.
-
But I cannot predict what you
perceive to be easier or more
-
difficult. And I'm
curious about it.
-
So let's see what you think.
-
Just keep an eye
on both of them.
-
Compare them, and then
tell me what you think
-
was more efficient or easier
to follow or understand.
-
OK.
-
I'll take this one
that's really pretty.
-
Example one.
-
It is practically
straight out of the book.
-
It appeared as an obsession
in several final exams
-
with little variations.
-
The constraint was a
little, pretty one.
-
It's a linear constraint that
you have on the variables.
-
-
The g function I was talking
about, the marriage constraint,
-
is x plus y.
-
And this is the c, little
c we were talking about.
-
-
So how do I know that there
exists an x0, y0 extreme?
-
How do I know there
is an x0, y0 extreme?
-
-
I need to look baffled.
-
How do I look?
-
-
I don't know.
-
I'm just thinking, well,
maybe I can find it.
-
And once it verifies
all the conditions
-
of Lagrange's theorem, that
I know I'm in business--
-
and I would compute everything,
and compare the values,
-
and get my max and my min.
-
So what do I need
to do in step one?
-
Step one.
-
Oh, my god.
-
You guys, remind me,
because I forgot.
-
I'm just pretending, of course.
-
But I want to see if you
were able to remember.
-
The two gradients of
f and g, respectively,
-
have to be proportional.
-
That's kind of the idea.
-
And the proportionality
factor is lambda.
-
So I do f sub x
equals lambda g sub x.
-
f sub y equals lambda g sub y,
assuming that the gradient of g
-
is non-zero at that point where
I'm looking and assuming that g
-
of xy equals-- guys, I'm-- well,
OK, I'm going to write it now.
-
But then I have to say
who these guys are,
-
because that's the
important thing.
-
And this is where
I need your help.
-
-
So you tell me.
-
Who is this fellow, f sub x?
-
STUDENT: Negative 2x.
-
PROFESSOR: Minus
2x equals lambda.
-
Lambda.
-
Mr. Lambda is important.
-
I'm going to put it in red.
-
Know why I'm putting him
in red-- because he needs
-
to just jump into my eyes.
-
Maybe I can
eliminate the lambda.
-
This is the general philosophy.
-
Maybe to solve the system,
I can eliminate the lambda
-
between the equations somehow.
-
-
How about Mr. g sub x?
-
g sub x is 1, so
it's a blessing.
-
I shouldn't write
times 1, but I am silly
-
and you know me by now.
-
So I'm going to keep going.
-
And I say, minus two
more is the same way.
-
Mr. Lambda in red very
happy to be there.
-
And times--
-
STUDENT: 1 again.
-
PROFESSOR: 1 again.
-
Thank god.
-
And then this easy condition,
that translates as x plus y
-
is 1.
-
And now what do we do?
-
Now we start staring at the
system, and we see patterns.
-
And we think, what
would be the easiest way
-
to deal with these patterns?
-
We see a pattern like
x plus y is known.
-
And if we were to sum
up the two equations,
-
like summing up the left-hand
side and right-hand side,
-
x plus y would be included as
in something in there as a unit.
-
So I'm just trying to
be creative and say,
-
there is no unique
way to solve it.
-
You can solve it in many ways.
-
But the easiest way
that comes to mind
-
is like, add up the left-hand
side and the right-hand side.
-
How much is that, the
lambda plus lambda?
-
STUDENT: 2 lambda.
-
PROFESSOR: 2 lambda, right?
-
And the x plus y is 1, because
God provided this to you.
-
You cannot change this.
-
OK?
-
It's an axiom.
-
So you replace it here.
-
1.
-
And you say, OK, I divide by 2.
-
Whatever.
-
Lambda has to be minus 1.
-
So if lambda is minus 1, do
I have other possibilities?
-
So first thing, when you
look at this algorithm,
-
you say, well, I know
what I have to do,
-
but are there any
other possibilities?
-
And then you say no,
that's the only one.
-
For lambda equals minus 1,
fortunately, you get what?
-
x equals 1/2.
-
Unless-- give it a name,
because this variable
-
means an arbitrary variable.
-
It's 0.
-
It's not-- and for
the same, you get
-
y0 equals 1/2 in the same way.
-
And then you say, OK,
for I know x0 y0 are now,
-
that's the only extremum that
I'm having to look at for now.
-
What is the 0?
-
So I'm going to
go ahead and say,
-
the point will be P0, 1/2, 1/2.
-
And then I plug in, and I
say, 1 minus 1/4 minus 1/4
-
is again 1/2.
-
-
When you compute problems,
when you computationally
-
solve problems,
many times you're
-
going to see that you
make algebra mistakes.
-
If you think I
don't make them, you
-
have proof that I make
them myself sometimes.
-
What is the best way
to protect yourself?
-
-
When you get numerical
answers a little bit,
-
see if they make sense.
-
Does that make sense?
-
Yes.
-
Is the sum 1?
-
Yes.
-
A little bit of double-checking
with your constraints,
-
your original data,
it looks good.
-
All right.
-
So the question
here is-- right now
-
the question is, are we done?
-
The answer is no,
we haven't quite
-
looked at what happens
with the constraint g,
-
because c-- oh, I forgot
to tell you that the book--
-
if you look in the book--
that's why you should have
-
the book in electronic format,
so you can read it in Kindle.
-
Example one had the
additional requirement
-
that x and y are positive.
-
Is such a requirement
natural in applications
-
of calculus, because this is
Calculus 3 with applications.
-
Can you give me an example
where x and y, being positive,
-
would be a must?
-
STUDENT: When they're
both distances?
-
PROFESSOR: Distances
or some physical things
-
that are measurable.
-
Lengths, widths, the
girth around an object,
-
some positive numbers that-- OK.
-
All right.
-
So we will see an example
involving dimensions
-
of a box and volume of a
box, where, of course, x,
-
y, z will be the
length, the width,
-
and the height of the box.
-
So that would come naturally
as x, y, z positive.
-
Next we are going to do that.
-
Now, what's going to happen
for this kind of constraint?
-
So I want to see if x and y are
positive but at the same time,
-
they are married to
Px plus y equals 1,
-
I do not have just
all the possibilities.
-
I have to have in mind
their picture as a couple.
-
x plus y, as a couple,
must be 1, meaning you get
-
the segment, this segment.
-
Are you guys with me?
-
Why don't I expand
to the whole line?
-
I say, I want to expand
to the whole line, which
-
would be stupid.
-
Why would it be stupid?
-
I would get y equals
something negative here.
-
And if I expand in the other
direction, x would be negative.
-
So it's not a good thing.
-
So the only thing I
have is the segment,
-
which has two endpoints.
-
Those two endpoints
are boo-boos.
-
-
The endpoints can give
you extrema as well.
-
We talked about it last time.
-
So every time you do
this, you're fine,
-
but you have to compare the
results against the extrema.
-
These are artificial cuts.
-
In what sense artificial?
-
In the sense that you
don't let the whole thing
-
evolve over the
whole real domain.
-
Once you artificially
cut something--
-
let me give you another example.
-
Don't put this in
the notes, because I
-
don't want it to confuse you.
-
You have some natural, so-called
relative max and minima here,
-
right?
-
That's a relative min, that's
a relative max, and so on.
-
If I make an artificial
cut anywhere-- let's
-
say this is not going
to be a minimum anymore.
-
I make an artificial cut here,
I make an artificial cut here.
-
These endpoints will
generate other possibilities
-
for my absolute max
and absolute min.
-
So those extrema are
extremely important.
-
I have one.
-
What is this guy's-- 1, 0.
-
Am I right?
-
And this is 0,1.
-
So I have to look
at the possibility.
-
When it's 1 by 1, it
goes-- say it again.
-
1,0.
-
And x2y2 was hmm?
-
0,1.
-
And of course, both
of them satisfy that.
-
In this case, f of 1, 0 has
to be evaluated as well.
-
That's going to be 1 minus
1 squared minus 0 equals 0.
-
And by the symmetry
of this polynomial,
-
you are going to have the
same answer, 0, in both cases.
-
You're going to draw the table.
-
And this is the perfect
place for the table.
-
Perfect place in the
sense that you have x, y
-
and you have-- who are your
notable, noticeable guys?
-
1/2, 1/2.
-
Who said 1,0?
-
And 0, 1.
-
And who was the zz
was the 1/2 here.
-
And here was 0, and here was 0.
-
And I'm going to start
making faces and drawing.
-
-
Did I get the answer?
-
Did I solve this
problem at home?
-
Yes, I did.
-
And I got the same answer.
-
All right.
-
So this is max.
-
This is min.
-
This is mean, the same.
-
So both of these are what?
-
Absolute minima.
-
And these are the
absolute extrema
-
for this problem
with constraint.
-
I'm going to go ahead
and erase and say,
-
remember in the
eyes of your mind
-
how much work it was to do this?
-
And I'm going to apply
the other method.
-
So how much space?
-
So we needed one
board largely written.
-
You want to go to follow
the steps one and two.
-
Should I erase that?
-
STUDENT: No.
-
-
PROFESSOR: You
are my note-taker.
-
Of course I will listen to you.
-
And then let's see what method
number two I had in mind
-
is the one from last time.
-
So this is a what?
-
It's a review of-- what
is the section time?
-
11.7.
-
And I'm going to make a face,
happy that I can have yet
-
another application for you.
-
When this problem appeared on
the final at least five times
-
in the last 10 finals or
more, different instructors
-
viewed it differently.
-
Practically, the
general instruction
-
given to the students--
solve it anyway you
-
find it easier for you.
-
Just don't make mistakes.
-
So we did not
encourage instructors
-
to say, do this by
Lagrange multipliers,
-
or do this by-- no, no, no, no.
-
Whatever is easier
for the student.
-
So what did we do last
time about the constraint?
-
Since x and y are
married, y depend on x.
-
So y is 1 times x.
-
And we say, this
is my guy that I
-
have to plug in
into the function,
-
into the original function.
-
And then f would
not be a function
-
of two independent
variables anymore.
-
But it's going to become a
function of one variable.
-
Thank god it's not hard.
-
It's no hard
because in this way,
-
you have just to pull
out the y1 minus x,
-
and square it, and
do the algebra.
-
So 1 minus x squared.
-
And I'm going to do
this really quickly.
-
Minus 1 minus x
squared and plus 2x.
-
-
And OK.
-
So we say, all right, all
right, so 1 and minus 1 go away.
-
-
It make our life easier,
because I have minus 2x squared
-
plus So of course,
I could do it fast,
-
but the whole idea
is not to amaze you
-
with my capability of working
fast, but be able to follow.
-
So you have minus what?
-
So tell me.
-
You can pull out a minus 2x.
-
And you get x.
-
And a minus 1.
-
-
And what is special about that?
-
Well, do I really
need to do that?
-
That's the question.
-
Could I have stopped here?
-
Is this the point
of factoring out?
-
Not really because factoring
out is not going to help you.
-
What I want is to chase
after Mr. f prime of x
-
and solve the critical point
equation f prime of x equals 0.
-
Right?
-
I need to find that x0 that will
satisfy f prime of x equals 0.
-
What do I get?
-
I get minus 4x plus 2 equals 0.
-
And I see I'm already relieved.
-
The moment I saw that, I
felt that I'm doing this
-
the right way, because I had
the previous method that led me
-
to a 1/2 that Alex provided for
somebody for the first time.
-
So now I feel I'm going
to get the same thing.
-
Let's see how much faster
or how much slower.
-
Why 0 corresponding
to it will be 1/2?
-
Because 1/2 plus 1/2 is a 1.
-
So what do I do?
-
Just as before, I start
my table and I say,
-
x and y must be 1/2 and 1/2
to give me the critical point
-
in the middle.
-
And I'm going to
get a 1/2 for that.
-
And I don't yet.
-
I pretend.
-
I don't know that's
gonna be a maximum.
-
What other points will provide
the books, the so-called--
-
STUDENT: Endpoints.
-
PROFESSOR: The endpoints.
-
And for those endpoints,
I keep in mind
-
that x and y,
again, are positive.
-
I should keep this picture in
mind, because if I don't, well,
-
it's not going to be very good.
-
So x is not allowed to move.
-
See, x has limited freedom
from the constraint.
-
So he's not allowed
to leave this segment.
-
x is going to be
between 0 and 1.
-
So for the endpoint x equals
0-- will provide me with y
-
equals 1.
-
And I'll put it in the
table, and I'll say,
-
when x equals 0 and y
equals 1-- and in that case,
-
I plug back in here and I get 0.
-
And again, for the
same type of-- I mean,
-
the other endpoint, I
get 1,0, and I get 0.
-
And it's the same thing.
-
I got the same thing
through another method.
-
This is the max, and
these are the mins.
-
And one of my students asked me
in my office hour-- by the way,
-
if you cannot make it to
Tuesday's 3:00 to 5:00,
-
you can come today.
-
At 2:00 after we are done,
I'm going to be in my office
-
as well.
-
So just.
-
So I have Tuesdays and Thursdays
after class, right after class.
-
Now, no matter what, if you get
the same answer, what if you
-
forget about one of the values?
-
Like, this student asked me,
what if I got the right maximum
-
and I got the right minimum, and
I say those are your extrema,
-
and I don't prove, mind you,
both points when it happens,
-
only one?
-
I don't know.
-
It's different from a
problem to the other.
-
Maybe I'm subtracting
some credit.
-
But you get most of the
partial credit in that case.
-
There will be many
values in which
-
you get the same altitude.
-
This is the altitude.
-
My z.
-
-
Do you have questions of that?
-
OK.
-
Now it's my turn
to make you vote.
-
And if you cannot
vote, you abstain.
-
Which one was easier?
-
The first method, the
Lagrange multipliers?
-
Or the second one, the-- how
should I call the second one,
-
the--
-
STUDENT: Integration.
-
PROFESSOR: The ray substitution
method then derivation,
-
count one type method?
-
So who is for-- OK.
-
You got this on the
midterm, say, or final.
-
How many of you would feel the
first method would be easier
-
to employ?
-
STUDENT: The second.
-
PROFESSOR: And how many of
you think the second method
-
is easier to employ?
-
And how many people
say that they
-
are equally long, or short,
or how many people abstain?
-
STUDENT: I would say it
depends on the problem.
-
PROFESSOR: Yeah, absolutely.
-
But I'm talking about
this particular one,
-
because I'm curious.
-
STUDENT: Oh, on this one.
-
Oh, OK.
-
PROFESSOR: I'm going to go
on and do another problem.
-
And for that, also, I will ask.
-
-
with other problems, it
may be that it's easier
-
to solve the system for
the Lagrange multipliers
-
than it is to pull out the y
explicitly from the constraint
-
and put it back.
-
-
What else have I
prepared for you?
-
I had cooked up something.
-
-
I had cooked up
some extra credit.
-
But I don't know
if you have time.
-
But write it anyway.
-
So please write
down, for one point
-
extra credit for
the next seven days,
-
read and summarize both
of the following methods--
-
Lagrange multipliers with
one parameter lambda,
-
which is exactly the
same I taught you.
-
Same I taught.
-
And one that is not required
for the examinations, which
-
is Lagrange multipliers
with two parameters.
-
-
And that is a big
headache when you
-
do that, because you
have two parameters.
-
Let's call them lambda and mu.
-
I don't know what to call them.
-
When you have that kind
of method, it's longer.
-
So it may take you several
pages of computation
-
to get to the lambdas and to
the extrema and everything.
-
But I would like you to
at least read the theorem
-
and write down a short
paragraph about one of these.
-
So both of them are one point.
-
Both of them are one point
extra credit at the end.
-
STUDENT: Together or each?
-
PROFESSOR: Yes, sir?
-
STUDENT: One point each
or one point together?
-
PROFESSOR: No, one
both, together.
-
I'm sorry.
-
Because there will be other
chances to get extra credit.
-
And I'm cooking up something
I didn't say on the syllabus,
-
like a brownie point
or cake or something.
-
At the end of the
class, I would like
-
you to write me a statement
of two pages on how
-
you think Calculus 3
relates to your major.
-
And one question from
a previous student
-
was, I've changed
my major four times.
-
Which one shall I pick?
-
I said, whichever
you are in right now.
-
How does that relate?
-
How is Calculus 3
relevant to your major?
-
Give me some
examples and how you
-
think functions of two
variables or three variables--
-
STUDENT: What's this?
-
PROFESSOR: Up here
in your main major.
-
STUDENT: So it's a two
parameter question?
-
Like, would there be any
question regarding that?
-
PROFESSOR: No, nothing.
-
Not in the homework.
-
We don't cover that, we
don't do that in the test.
-
Most instructors
don't even mention it,
-
but I said, mm, you
are our students,
-
so I want to let you do
a little bit of research.
-
It's about a page and
a half of reading.
-
Individual study.
-
STUDENT: Is that in the book?
-
PROFESSOR: It is in the book.
-
So individual study.
-
One page or one page and a half.
-
Something like that.
-
Maybe less.
-
OK.
-
-
One other one that I cooked
up-- it's not in the book.
-
But I liked it because it sounds
like a real-life application.
-
It is a real-life application.
-
And I was talking
to the mailman.
-
And he was saying, I wonder
how-- because a guy, poor guy,
-
was carrying these
Priority boxes.
-
And he said, I wonder
how they optimize?
-
When they say "flat
rate," how do they
-
come up with those dimensions?
-
And it's an
optimization problem,
-
and there are many like
that in the real world.
-
But for my case,
I would say, let's
-
assume that somebody says,
the sum of the lengths
-
plus widths plus height is
constrained to be some number.
-
x plus y plus z equals
the maximum possible.
-
Could be 50 inches.
-
But instead of 50
inches-- because I
-
don't want to work with
that kind of numbers,
-
I'm too lazy-- I put x
plus y plus equals 1.
-
That's my constraint, g.
-
-
I would like to
maximize the volume.
-
Say it again, Magdalena.
-
What is the problem?
-
What's your problem?
-
My problem is example three.
-
Maximize the volume of a box
of length, height, and width x,
-
y, z, just to make our
life easier in a way
-
that the girth cross
the-- well, OK.
-
Let me make this interesting.
-
The sum of the
dimensions equals 1.
-
And where can you
find this problem?
-
Well, this problem can be
found in several resources.
-
We haven't dealt very
much with functions
-
of three variables, x, y, z.
-
But the procedure
is exactly the same.
-
I stole that from
a library online
-
that's called Paul's
Online Calculus Notes.
-
And imagine that
the same thing I
-
taught you would be applied to
functions of three variables.
-
Tell me who the volume will be.
-
The volume would be a
function of three variables.
-
Let's call it f, which is what?
-
Who is telling me what?
-
STUDENT: x times y.
-
PROFESSOR: x times y.
-
Thanks.
-
And are we happy about it?
-
Ah, it's a beautiful function.
-
It's not going to give you
too much of a headache.
-
-
I would like you to cook up
step one and step two for me
-
by the Lagrange
multipliers I specify.
-
-
For functions of
three variables.
-
-
Maximize and minimize.
-
-
Yeah.
-
-
OK.
-
So the gradients are not
going to be in R2 anymore.
-
They will be in R3.
-
And so what?
-
It doesn't matter.
-
Step one.
-
-
Say it again, Magdalena.
-
What do you mean?
-
I mean that when you're going
to have something like that,
-
the system for nabla f of
x, y, z at the point x0, y0,
-
z0 will be lambda times
nabla g of x at x0, y0 is 0,
-
where both nablas are in R3.
-
Right?
-
They will be f sub x, f sub
y, f sub z angular brackets.
-
So instead of having just
two equations in the system,
-
you're going to have
three equations.
-
That's the only big difference.
-
Big deal.
-
Not a big deal.
-
So you tell me what
I'm going to write.
-
So I'm going to write f sub
x equals lambda g sub x.
-
f sub y equals lambda g sub y.
-
f sub z equals lambda g sub z.
-
Thank god I don't have
more than three variables.
-
Now we-- in fact, it's
how do you think engineers
-
solve this kind of system?
-
Do they do this by hand?
-
No.
-
Life is complicated.
-
When you do Lagrange multipliers
on a thermodynamical problem
-
or mechanics problem,
physics problem,
-
you have really ugly
data that are programs
-
based on Lagrange multipliers.
-
You can have a
Lagrange multiplier
-
of seven different parameters,
including pressure, time,
-
and temperature.
-
And it's really horrible.
-
And you don't do that by hand.
-
That's why we have to be
thankful to technology
-
and the software, the
scientific software methods.
-
You can do that in MATLAB, you
can do that in Mathematica.
-
MATLAB is mostly for engineers.
-
There are programs written
especially for MATLAB
-
to solve the problem of
Lagrange multipliers.
-
Now, this has not complete.
-
We are missing the most
important, the marriage thing,
-
the g of x, y, z constraint.
-
Now there are three
in the picture.
-
I don't know what that means.
x plus y plus z equals 1.
-
-
So if and only if, who's going
to tell me what those will be?
-
Are they going to be hard?
-
No.
-
It's a real-life problem, but
it's not a hard problem. f of x
-
will be yz equals lambda.
-
Who is g sub x?
-
STUDENT: 1.
-
PROFESSOR: 1.
-
Thank god.
-
So it's fine.
-
It's not that.
-
F sub y would be
xz equals lambda.
-
f sub z is xy equals lambda.
-
Ah, there is a lot
of symmetry in that.
-
I have some thinking to do.
-
Well, I'm a scientist.
-
I have to take into account
all the possibilities.
-
If I lose one, I'm dead
meat, because that one
-
may be essential.
-
So if I were a computer,
I would branch out
-
all the possibilities
in a certain order.
-
But I'm not a computer.
-
But I have to think in
the same organized way
-
to exhaust all the
possibilities for that.
-
And for that matter, I
have to pay attention.
-
So I have x plus
y plus z equals 1.
-
OK.
-
I'm going to give you
about-- we have time?
-
Yes.
-
I'll give you two
minutes to think
-
how to solve-- how
does one solve that?
-
How does one solve it?
-
Think how you would grab.
-
Where would you grab
the problem from?
-
But think it for yourself, and
then I'm gone for two minutes.
-
And then I'm going to
discuss things out loud,
-
and I'll share with
you how I did it.
-
STUDENT: It could
be 1, 1 minus 1.
-
PROFESSOR: You are
like an engineer.
-
You already see, oh, maybe
I could have some equality
-
between the coordinate.
-
We have to do it in
a mathematical way.
-
All right?
-
So would it help me if I
subtracted the second equation
-
from the first equation?
-
What kind of
information would I get?
-
STUDENT: But that
can be your ratio.
-
STUDENT: We can divide better.
-
-
PROFESSOR: I can divide.
-
That's another possibility.
-
I can divide and
do y/x equals 1.
-
And that would give
you x equals y.
-
And then you plug it back.
-
And then you say, wait a minute.
-
If x equals y, then
x times x is lambda.
-
So lambda would be x squared.
-
So then we plug it in here.
-
And we go, x plus x equals 2x.
-
And then we see what else we
can find that information.
-
As you can see, there is no
unique way of doing that.
-
But what's unique
should be our answer.
-
No matter how I do it, I should
overlap with Nitish's method.
-
At some point, I should
get the possibility
-
that x and y are the same.
-
If I don't, that means
I'm doing something wrong.
-
So the way I approach this
problem-- OK, one observation,
-
I could subtract the second
from the first, where
-
I would subtract the
third from the second.
-
Or I could subtract the
second from the first
-
and analyze all
the possibilities.
-
Let's do only one
and then by symmetry,
-
because this is a
symmetric problem.
-
By symmetry, I'm going to
see all the other problems.
-
So how do you think in symmetry?
-
x and y and z have-- it's a
democratic world for them.
-
They have the same roles.
-
So at some point when you
got some solutions for x, y,
-
z in a certain way,
you may swap them.
-
You may change the
rules of x, y, z,
-
and get all the solutions.
-
So the way I did it was I took
first xz minus yz equals 0.
-
-
But then let's interpret what
this-- and a mathematician
-
will go either by if and
only-- if/or it implies.
-
I don't know if
anybody taught you.
-
Depends where
you're coming from,
-
because different
schools, different states,
-
different customs
for this differently.
-
But in professional mathematics,
one should go with if and only
-
if, or implication,
x minus yz equals 0.
-
-
And then what
implication do I have?
-
Now I don't have an implication.
-
I have it in the sense
that I have either/or.
-
So this will go, like in
computer science, either/or.
-
Either-- I do the branching--
x equals y, or z equals 0.
-
And I have to study
these cases separately.
-
You see?
-
It's not so obvious.
-
Let me take this one, because
it's closer in my area
-
on [INAUDIBLE].
-
It doesn't matter in
which order I start.
-
For z equals 0, if I plug in
z equals 0, what do I get?
-
Lambda equals 0, right?
-
-
But if lambda equals 0, I
get another ramification.
-
So you are going to say,
oh, I'm getting a headache.
-
Not yet.
-
So lambda equals 0 will again
lead you to two possibilities.
-
Either x equals 0 or--
-
STUDENT: y equals 0.
-
PROFESSOR: --y equals 0.
-
-
Let's take the first one.
-
Like a computer,
just like a computer,
-
computer will say,
if-- so I'm here.
-
If 0 is 0 and x was
0, what would y be?
-
y will be 1.
-
That is the only case I got.
-
And I make a smile, because
why do I make a smile now?
-
Because I got all three of
them, and I can start my table
-
that's a pink table.
-
And here I have x, y,
z significant values.
-
Everything else doesn't matter.
-
And this is z, which was the
volume, which was x, y, z.
-
Was it, guys?
-
So I have to compare volumes
for this thinking box.
-
Right?
-
OK.
-
Now, in this case, I have 0, x.
-
y is 1.
-
z is 0.
-
The volume will be--
and do I have a box?
-
No, I don't have a box.
-
I make a face like that.
-
But the value is
still there to put.
-
As a mathematician, I
have to record everything.
-
STUDENT: Do you have to
put this on the exam?
-
Because it doesn't make sense.
-
This would not--
-
PROFESSOR: No.
-
No.
-
Because I haven't said,
if the box cannot be used.
-
I didn't say I
would use it or not.
-
So the volume 0 is a possible
value for the function.
-
And that will give
us the minimum.
-
So what do we-- I expect
you to say in the exam,
-
I have the absolute minimum.
-
One of the points-- I'm
going to have more points
-
when I have minima.
-
OK.
-
And the other case.
-
I don't want to get
distracted. y is 0.
-
So I get x equals 1.
-
Are you guys with me?
-
From here and here
and here, I get
-
x equals 1, because the sum
of all three of them will be,
-
again, 1.
-
So I have another pair.
-
0, 0.
-
STUDENT: Wouldn't it be 1, 0?
-
PROFESSOR: 1, 0.
-
1, 0, 0.
-
And the volume will be the same.
-
And another absolute minima.
-
Remember that everything
is positive-- the x, y, z,
-
and the [INAUDIBLE].
-
I keep going.
-
And I say, how do I get-- I
have the feeling I'm going
-
to get 0, 0, 1 at some point.
-
But how am I going
to get this thing?
-
I'm going to get
to it naturally.
-
So I should never anticipate.
-
-
The other case
will give it to me.
-
OK?
-
So let's see.
-
When x equals y, I
didn't say anything.
-
When x equals y, I have
to see what happens.
-
And I got here again two cases.
-
Either x equals-- either,
Magdalena, either x
-
equals y equals 0, or x
equals y equals non-zero.
-
So I'm a robot.
-
I'm an android.
-
I don't let any logical
piece escape me.
-
Everything goes in
the right place.
-
When x equals y equals
0, the only possibility I
-
have is z to the 1.
-
And I make another face.
-
I'm why?
-
Happy that I'm at the end.
-
But then I realize
that it is, of course,
-
not what I hoped for.
-
It's another minimum.
-
So I have minima
0 for the volume
-
attained at all these three
possibilities, all the three
-
points.
-
-
And then what?
-
Then finally something
more interesting.
-
Finally.
-
x equals y different from 0.
-
What am I doing to
do with that case?
-
Of course, you can
do this in many ways.
-
But if you want to know what I
did, just don't laugh too hard.
-
I said, look, I'm
changing everything
-
in the original thing.
-
I'll take it aside, and
I'll plug in and see
-
what the system becomes.
-
So we'll assume x equals
y different from 0,
-
and plug it back in the system.
-
In that case, xy equals
lambda will become x squared
-
equals lambda, right?
-
Mr. x plus y plus
z equals 1 would
-
become x plus x plus z, which
is 2x plus z, which is 1.
-
-
And finally, these
two equations,
-
since x equals y are one and
the same, they become one.
-
xz equals lambda.
-
And I stare at this guy.
-
And somebody tell
me, can I solve that?
-
Well, it's a system,
not a linear system.
-
But it's a system
of three variables.
-
Three equations-- I'm sorry--
with three unknowns-- x, z,
-
and lambda.
-
So it should be easy
for me to solve it.
-
How did I solve it?
-
I got-- it's a little bit funny.
-
I got x equals lambda over z.
-
And then I went-- but let
me square the whole thing.
-
And I'm going to get--
why do I square it?
-
Because I want to compare
it to what I have here.
-
If I compare, I go, if
and only if x squared
-
equals lambda squared
over z squared.
-
But Mr. x squared is
known as being lambda.
-
So I will replace
him. x squared is
-
lambda from the first equation.
-
So I get lambda equals lambda
squared over z squared.
-
So I got that-- what did I get?
-
Nitish, tell me.
-
Lambda equals?
-
STUDENT: x squared.
-
PROFESSOR: z squared.
-
STUDENT: Lambda is
equal to z squared.
-
PROFESSOR: So if and only
if lambda equals z squared.
-
But lambda was x
squared as well.
-
So lambda was what?
-
Lambda was z squared,
and lambda was x squared.
-
And it implies that x equals z.
-
x is equal to z.
-
But it's also equal to y.
-
Alex jump on me.
-
Why would that be?
-
STUDENT: Because
you just said that--
-
PROFESSOR: Because x was
y from the assumption.
-
So equal to y.
-
So this is the beautiful thing,
where all the three dimensions
-
are the same.
-
-
So what do we know that
thingie-- x equals z equals y?
-
STUDENT: It's a box.
-
PROFESSOR: It's a box of a what?
-
STUDENT: It's a square.
-
STUDENT: Square.
-
PROFESSOR: It's a--
-
ALL STUDENTS: Cube.
-
PROFESSOR: Cube.
-
OK.
-
So for the cube--
-
STUDENT: Square box.
-
PROFESSOR: We get--
for z, they were
-
stingy about the
dimensions we can have.
-
So they said, x plus y plus
z should be, at most, 1.
-
But we managed to maximize
the volume by the cube.
-
The cube is the only one
that maximizes the volume.
-
How do I get it back?
-
So I get it back by saying,
x plus y plus z equals 1.
-
So the only possibility that
comes out from here is that--
-
STUDENT: They're all 1/3.
-
PROFESSOR: That I
have 1/3, 1/3, 1/3.
-
And I have to take
this significant point.
-
This is the significant
point that I was praying for.
-
And the volume will be 1/27.
-
-
And I'm happy.
-
Why am I so happy?
-
Is this 1/27 the best I can get?
-
In this case, yes.
-
So I have the maximum.
-
Now, assume that
somebody would have--
-
STUDENT: That's a
really small box.
-
PROFESSOR: It's a small box.
-
Exactly.
-
I'm switching to
something, so assume--
-
I don't know why airlines
do that, but they do.
-
They say, the girth plus
the height will be this.
-
Girth meaning-- the
girth would be--
-
so this is the height
of your-- can I get?
-
Or this one?
-
No, that's yours.
-
Oh.
-
It's heavy.
-
You shouldn't make
me carry this.
-
OK.
-
So x plus y plus x
plus y is the girth.
-
And some airlines
are really weird.
-
I've dealt with at least
12 different airlines.
-
And the low-cost airlines that
I've dealt with in Europe,
-
they don't tell you what.
-
They say, maximum, 10-kilo max.
-
That's about 20 pounds.
-
And the girth plus the length
has to be a certain thing.
-
And others say just the--
some of the three dimensions
-
should be something like that.
-
Whatever they give you.
-
So I know you don't think
in centimeters usually.
-
But imagine that
somebody gives you
-
the sum of the three
dimensions of your check-in bag
-
would be 100.
-
That is horrible.
-
What would be the maximum
volume in that case?
-
STUDENT: It would all be
33 and 1/3 centimeters.
-
PROFESSOR: Huh?
-
STUDENT: It would all be
33 and 1/3 centimeters.
-
PROFESSOR: You mean?
-
STUDENT: They'll all be 33 and
1/3 centimeters, x, y, and z.
-
PROFESSOR: Not the sum. x
plus y plus z would be 100.
-
STUDENT: So each one of them?
-
PROFESSOR: And then you
have 33.33 whatever.
-
And then you cube that,
and you get the volume.
-
Now, would that be practical?
-
STUDENT: No.
-
PROFESSOR: Why not?
-
-
STUDENT: It doesn't fit.
-
PROFESSOR: It doesn't the
head bin and whatever.
-
So we try to-- because
the head bin is already
-
sort of flattened out, we
have the flattened ones.
-
But in any case,
it's a hassle just
-
having to deal with
this kind of constraint.
-
And when you come back
to the United States,
-
you really feel-- I don't know
if you have this experience.
-
The problem is not in
between continents.
-
-
You have plenty of-- you
can check in a baggage.
-
But if you don't, which I
don't, because I'm really weird.
-
I get a big carry-on,
and I can fit that.
-
And I'm very happy.
-
I have everything I need for
three weeks to one month.
-
But if you deal with low-cost
airlines, on that kind of 70
-
euro or something between London
and Milan, or Paris, or London
-
and Athens, or something,
and you pay that little, they
-
have all sorts of weird
constraints like this one.
-
x plus y plus z has to
be no more than that.
-
And the weight should be
no more than 20 pounds.
-
And I'll see how
you deal with that.
-
It's not easy.
-
So yes, we complain about
American airlines all the time,
-
but compared to those airlines,
we are really spoiled.
-
In the ticket price,
we are paying,
-
let's say, $300 from
here to Memphis,
-
we have a lot of goodies
includes that we may not always
-
appreciate.
-
I'm not working for
American Airlines.
-
Actually, I prefer
Southwest a lot
-
by the way they treat
us customers and so on.
-
But I'm saying,
think of restrictions
-
when it comes to
volume and weight,
-
because they represent something
in real-life applications.
-
Yes, sir.
-
STUDENT: I have a question
about the previous problem.
-
I found the 1/3 just
by finding the ratio--
-
the first and the second and
then the second and the third.
-
PROFESSOR: That's how Nitish--
-
STUDENT: Yeah, that's
how I did it as well.
-
PROFESSOR: You were
napping a little bit.
-
But yeah.
-
But then you woke up.
-
[LAUGHTER]
-
While you were napping, he goes,
divide by the first equation
-
by the second one,
and you get 1.
-
x/y is 1.
-
And so you get the solution
of having all of them equal,
-
all three.
-
STUDENT: Yeah.
-
Just because I did that
out, and then I was like,
-
oh, it's y is equal to x,
y is equal to z, and then
-
just change it all to y.
-
PROFESSOR: Right.
-
So how do you think I'm going
to proceed about your exams?
-
Do I care?
-
No.
-
As long as you get the right
answer, the same answer,
-
I don't care which
method you were using.
-
The problem for me comes where
you have had the right idea.
-
You messed up in the
middle of the algebra,
-
and you gave me the
wrong algebraic solution.
-
That's where I have to ponder
how much partial credit
-
I want to give you
or not give you.
-
But I'm trying to be fair,
in those cases, to everybody.
-
I wanted to tell you-- I
don't know if you realize,
-
but I stole from you every
Tuesday about seven minutes
-
from your break.
-
It should be a little
bit cumulative.
-
I'm going to give you
back those minutes
-
right now, hoping that
those seven minutes,
-
you're going to use them doing
something useful for yourself.
-
-
At the same time, I'm
waiting for your questions
-
either now, either here,
or in my office upstairs.
-
And I know many of you
solved a lot of the homework.
-
I'm proud of you.
-
Some of you did not.
-
Some of you still struggle.
-
I'm there to help you.
-
Is it too early-- I
mean, somebody asked me,
-
if I read ahead Chapter 12,
can I have the homework early?
-
Is it too early?
-
I don't know what to do.
-
I mean, I feel it's
too early to give you
-
Chapter 12 [INAUDIBLE]
and Chapter
-
12 problems ahead of time.
-
But if you feel it's OK, I can
send you the homework next.
-
Yeah?
-
All right.
-
Whatever you want.
-
We will start
Chapter 12 next week.
-
So I extended the deadline
for Chapter 11 already,
-
and I can go ahead and start
the homework for Chapter 12
-
already.
-
And keep it for a month or so.
-
I feel that as long as you
don't procrastinate, it's OK.
-
STUDENT: I solved that one,
because I've seen it before.