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: 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.