TTU Math2450 Calculus3 Sec 11.7 and 11.8

Absolute extrema and Lagrange multipliers

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.