## TTU Math2450 Calculus3 Sec 11.7 and 11.8

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PROFESSOR: --here.
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I have excuse two
people for being sick.
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But I haven't
excused anybody else.
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You are not the complete group.
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I would like to take
attendance as soon as possible.
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Would you mind starting
an attendance sheet?
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STUDENT: Yeah, we
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PROFESSOR: Oh, you
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OK.
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I understand having
to struggle with snow,
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but you are expected
to come here.
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And I don't want to punish
the people who don't make it.
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I want to reward the people
who make it every time.
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That's the principle behind
perfect attendance for this.
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All right.
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Today we are going to
cover something new.
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It is new and it's not new.
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It's an extension of
the ideas in 11.7,
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which were finding extrema
of functions of the type
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z equals f of xy, and
classifying those.
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In 11.8, we provide a
very specific method.
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That's Lagrange multipliers.
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Of finding extrema,
you struggled.
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Well, you didn't
struggle, but it
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wasn't easy to find those
absolute extrema at every time.
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The Lagrange multipliers
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So practically, what
should we assume
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to know that the function of
two variables that we deal with
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is c1.
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Sometimes I assume it's smooth.
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What do we need?
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We need the derivatives.
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Derivatives.
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Derivative exist
and are continuous.
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I assume differentiability.
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OK?
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And what else do I assume?
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I assume that you have a
constraint that is also smooth.
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Let's say g of xy equals c.
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Do you remember?
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last time on Tuesday as well.
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So practically,
x and y are bound
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to be together by some sort of
agreement, contract, marriage.
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They depend on one another.
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They cannot leave
this constraint.
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And last time, I
really don't remember
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what problem I took last time.
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But we had something like, given
the function f of xy-- that
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was nice and smooth--
find the absolute maximum
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and the absolute
minimum of that function
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inside the-- or on
the closed disk.
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Remember that?
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disk, x squared
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plus y squared less
than or equal to 1.
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And we said, let's
find-- somebody gives you
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this very nice function.
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We found the critical
point inside.
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And we said, that's it.
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Relative max or relative min.
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Maybe we have more
depending on the function.
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What was crucial for us to
do-- to study the extrema that
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could come from the boundary.
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And in order for them to
come from the boundary,
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we played this little game.
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We took the boundary.
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We said, that's the circle.
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X squared plus y
squared equals 1.
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We pulled out the
y in terms of x
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and brought it back in
the original expression,
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z equals f of xy.
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Since y would depend on x as y
squared is 1 minus x squared,
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we plugged in and we got
a function of x only.
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For that function of x only
on the boundary, we said,
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we look for those critical
points for the function.
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It was a little bit of
time-consuming stuff.
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Critical points.
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That would give you
relative max or min
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for that function
on the boundary.
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Plus, we said, but that function
has n points in the domain,
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because the domain
would be for x
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between minus 1 and 1--
inclusively minus 1 and 1.
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So those minus 1 and 1's
as endpoints can also
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generate absolute max and min.
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So we made a table of
all the possible values,
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including all the critical
points and the values
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at the endpoints.
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We said, whoever's the
tallest guy over here's
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gonna be the maximum.
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Whoever's the smallest
one will be the minimum.
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And that was what the
philosophy was before.
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Now we have to find a
different method, which
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is providing the same solution,
but it's more systematic
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in approach and is
based on a result that
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was due to Lagrange, one
of the-- well, the fathers.
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The fathers were
Euler and Leibniz.
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of contributions
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to physics, mechanics
especially, and calculus.
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So he's also a father.
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As a father, he came up
with this beautiful theorem,
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that says, if you have
these conditions satisfied
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and if has-- if f has
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we know that has an extrema.
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At some point, P0 of x0 y0 along
the curve-- the boundary curve.
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Let's call this boundary curve
as script C. Do you understand?
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This is not an l.
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I don't know how to denote.
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Script C. Script C. How
do you draw a script C?
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Let's draw it like that.
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I'm not an l.
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OK?
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C from curve.
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Then there exists-- I
taught you the sign.
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There exists a lambda--
real number-- such
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What?
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of x0 y0 would be
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parallel of a proportionality
factor, lambda,
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the constraint function.
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So we have two
Musketeers here that
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the original function,
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the one you want to
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of the constraint function
as a function of x
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and y at the point.
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And we claim that
at that point, we
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have an extremum of some sort.
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Then something magical happens.
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There is a lambda-- there
is a proportionality
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So you say something magical
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will be in the same direction.
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And the proportionality factor
is this beautiful lambda.
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If Mr. g-- this is a tricky
thing you have to make sure
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happens.
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If Mr. g, let's say, is-- at
its 0y0 is different from 0.
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Because if it is equal to 0,
well, then it's gonna be crazy.
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We will have 0 equals 0 for
any lambda multiplication here.
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So that would complicate
things, and you would
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get something that's lost.
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So how do I view the procedure?
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How do I get the lambda?
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Once I grab this lambda,
I think I would be done.
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Because once I
grab the lambda, I
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could figure out who the x0,
y0 are from the equations.
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So I have this feeling I need a
procedure, I need an algorithm.
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Engineers are more
algorithmically oriented
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than us mathematicians.
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And this is what I appreciate
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They have a very
organized, systematic mind.
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So if I were to
write an algorithm,
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a procedure for the
method, I would say,
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assume that f and
g are nice to you.
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You don't say that.
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Don't write that.
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Now assume that f and g
satisfy Lagrange's theorem.
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Satisfy the conditions
of Lagrange's theorem.
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OK?
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The notion, by the way, has
nothing to do with Calc 3.
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But the notion of
Lagrangian and Hamiltonian
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are something you are
learning in engineering.
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And the Lagrangian is a
product of Mr. Lagrange.
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So he's done a lot for
science in general, not just
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for calculus, for mathematics,
for pure mathematics.
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OK.
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In that case, what
do you need to do?
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Step one.
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You need to recover that.
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So solve for x0, y0, and
lambda the following system.
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What does it mean the two
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to each other?
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This fella over here is
going to be what vector?
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f of xy-- f sub y.
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This gal over here will
be g sub x, g sub y.
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For them to be proportional,
you should have this,
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then-- f sub x equals g
sub x times the lambda.
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Right?
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And f sub y equals g
sub y times the lambda
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for the same lambda, your hero.
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So both coordinates have to be
multiplied by the same lambda
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to get you the other
partial velocities.
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STUDENT: And it has to be
evaluated at that point?
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PROFESSOR: Yes.
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So you're gonna solve for--
you're gonna solve this system,
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and you are going to
get-- and I'm sorry.
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With a constraint.
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And with the absolute constraint
that you have at g of xy
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equals c, because
these guys are married.
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They always are in
this relationship.
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And from all the
information of the system,
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you're going to get a-- not one,
maybe several values of lambda,
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you get values of lambda, x0,
y0, that satisfy the system.
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You have to satisfy all
the three constraints, all
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the three equations.
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I'm going to put them
in bullets, red bullets.
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You don't have colors, but I do.
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So.
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And then at all these points
that we found in step one,
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step two.
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Step two I'm going
to erase here.
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For all the points-- x0, y0,
and lambda 0-- you got step one.
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You have to evaluate
the f function.
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Evaluate f at those x0, y0,
lambda 0 we got 4 lambda 0.
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And get to compare values
in a table just like before.
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See all the points.
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All points will
give you an idea who
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is going to be the
absolute max, absolute min.
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And I'm just going to go ahead
and solve one typical example
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You know, it's not solved
in the book by both methods.
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But I'm thinking
since I'm teaching you
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how to apply the Lagrange
theorem today and do the step
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one, step two
procedure for Lagrange
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multipliers I'm going to solve
it with Lagrange multipliers
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first.
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And the same problem, I'm
going to solve it in the spirit
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that we have employed
last time in 11.7.
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And then I'm going
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which method is easier for you.
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And I'm really curious,
because of course, I
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can predict what theorems
I'm going to cover.
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And I can predict
the results I'm
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going to get in the exercises.
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But I cannot predict what you
perceive to be easier or more
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difficult. And I'm
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So let's see what you think.
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Just keep an eye
on both of them.
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Compare them, and then
tell me what you think
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was more efficient or easier
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OK.
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I'll take this one
that's really pretty.
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Example one.
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It is practically
straight out of the book.
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It appeared as an obsession
in several final exams
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with little variations.
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The constraint was a
little, pretty one.
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It's a linear constraint that
you have on the variables.
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The g function I was talking
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is x plus y.
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And this is the c, little
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So how do I know that there
exists an x0, y0 extreme?
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How do I know there
is an x0, y0 extreme?
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I need to look baffled.
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How do I look?
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I don't know.
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I'm just thinking, well,
maybe I can find it.
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And once it verifies
all the conditions
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of Lagrange's theorem, that
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and I would compute everything,
and compare the values,
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and get my max and my min.
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So what do I need
to do in step one?
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Step one.
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Oh, my god.
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You guys, remind me,
because I forgot.
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I'm just pretending, of course.
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But I want to see if you
were able to remember.
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f and g, respectively,
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have to be proportional.
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That's kind of the idea.
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And the proportionality
factor is lambda.
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So I do f sub x
equals lambda g sub x.
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f sub y equals lambda g sub y,
assuming that the gradient of g
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is non-zero at that point where
I'm looking and assuming that g
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of xy equals-- guys, I'm-- well,
OK, I'm going to write it now.
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But then I have to say
who these guys are,
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because that's the
important thing.
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And this is where
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So you tell me.
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Who is this fellow, f sub x?
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STUDENT: Negative 2x.
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PROFESSOR: Minus
2x equals lambda.
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Lambda.
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Mr. Lambda is important.
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I'm going to put it in red.
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Know why I'm putting him
in red-- because he needs
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to just jump into my eyes.
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Maybe I can
eliminate the lambda.
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This is the general philosophy.
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Maybe to solve the system,
I can eliminate the lambda
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between the equations somehow.
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How about Mr. g sub x?
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g sub x is 1, so
it's a blessing.
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I shouldn't write
times 1, but I am silly
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and you know me by now.
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So I'm going to keep going.
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And I say, minus two
more is the same way.
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Mr. Lambda in red very
happy to be there.
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And times--
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STUDENT: 1 again.
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PROFESSOR: 1 again.
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Thank god.
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And then this easy condition,
that translates as x plus y
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is 1.
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And now what do we do?
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Now we start staring at the
system, and we see patterns.
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And we think, what
would be the easiest way
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to deal with these patterns?
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We see a pattern like
x plus y is known.
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And if we were to sum
up the two equations,
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like summing up the left-hand
side and right-hand side,
• 18:41 - 18:46
x plus y would be included as
in something in there as a unit.
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So I'm just trying to
be creative and say,
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there is no unique
way to solve it.
• 18:52 - 18:54
You can solve it in many ways.
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But the easiest way
that comes to mind
• 18:57 - 19:02
is like, add up the left-hand
side and the right-hand side.
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How much is that, the
lambda plus lambda?
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STUDENT: 2 lambda.
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PROFESSOR: 2 lambda, right?
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And the x plus y is 1, because
God provided this to you.
• 19:13 - 19:16
You cannot change this.
• 19:16 - 19:17
OK?
• 19:17 - 19:19
It's an axiom.
• 19:19 - 19:21
So you replace it here.
• 19:21 - 19:23
1.
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And you say, OK, I divide by 2.
• 19:25 - 19:25
Whatever.
• 19:25 - 19:31
Lambda has to be minus 1.
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So if lambda is minus 1, do
I have other possibilities?
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So first thing, when you
look at this algorithm,
• 19:38 - 19:42
you say, well, I know
what I have to do,
• 19:42 - 19:45
but are there any
other possibilities?
• 19:45 - 19:48
And then you say no,
that's the only one.
• 19:48 - 19:54
For lambda equals minus 1,
fortunately, you get what?
• 19:54 - 19:58
x equals 1/2.
• 19:58 - 20:03
Unless-- give it a name,
because this variable
• 20:03 - 20:04
means an arbitrary variable.
• 20:04 - 20:05
It's 0.
• 20:05 - 20:08
It's not-- and for
the same, you get
• 20:08 - 20:13
y0 equals 1/2 in the same way.
• 20:13 - 20:17
And then you say, OK,
for I know x0 y0 are now,
• 20:17 - 20:24
that's the only extremum that
I'm having to look at for now.
• 20:24 - 20:25
What is the 0?
• 20:25 - 20:27
So I'm going to
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the point will be P0, 1/2, 1/2.
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And then I plug in, and I
say, 1 minus 1/4 minus 1/4
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is again 1/2.
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• 20:42 - 20:46
When you compute problems,
when you computationally
• 20:46 - 20:49
solve problems,
many times you're
• 20:49 - 20:52
going to see that you
make algebra mistakes.
• 20:52 - 20:55
If you think I
don't make them, you
• 20:55 - 20:58
have proof that I make
them myself sometimes.
• 20:58 - 21:00
What is the best way
to protect yourself?
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• 21:03 - 21:05
When you get numerical
• 21:05 - 21:07
see if they make sense.
• 21:07 - 21:08
Does that make sense?
• 21:08 - 21:09
Yes.
• 21:09 - 21:10
Is the sum 1?
• 21:10 - 21:11
Yes.
• 21:11 - 21:13
A little bit of double-checking
• 21:13 - 21:18
it looks good.
• 21:18 - 21:19
All right.
• 21:19 - 21:21
So the question
here is-- right now
• 21:21 - 21:25
the question is, are we done?
• 21:25 - 21:28
we haven't quite
• 21:28 - 21:32
looked at what happens
with the constraint g,
• 21:32 - 21:36
because c-- oh, I forgot
to tell you that the book--
• 21:36 - 21:39
if you look in the book--
that's why you should have
• 21:39 - 21:44
the book in electronic format,
so you can read it in Kindle.
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• 21:50 - 21:53
that x and y are positive.
• 21:53 - 21:56
Is such a requirement
natural in applications
• 21:56 - 22:00
of calculus, because this is
Calculus 3 with applications.
• 22:00 - 22:06
Can you give me an example
where x and y, being positive,
• 22:06 - 22:08
would be a must?
• 22:08 - 22:09
STUDENT: When they're
both distances?
• 22:09 - 22:13
PROFESSOR: Distances
or some physical things
• 22:13 - 22:15
that are measurable.
• 22:15 - 22:21
Lengths, widths, the
girth around an object,
• 22:21 - 22:26
some positive numbers that-- OK.
• 22:26 - 22:26
All right.
• 22:26 - 22:30
So we will see an example
involving dimensions
• 22:30 - 22:34
of a box and volume of a
box, where, of course, x,
• 22:34 - 22:36
y, z will be the
length, the width,
• 22:36 - 22:39
and the height of the box.
• 22:39 - 22:43
So that would come naturally
as x, y, z positive.
• 22:43 - 22:46
Next we are going to do that.
• 22:46 - 22:50
Now, what's going to happen
for this kind of constraint?
• 22:50 - 22:54
So I want to see if x and y are
positive but at the same time,
• 22:54 - 22:57
they are married to
Px plus y equals 1,
• 22:57 - 23:00
I do not have just
all the possibilities.
• 23:00 - 23:05
I have to have in mind
their picture as a couple.
• 23:05 - 23:11
x plus y, as a couple,
must be 1, meaning you get
• 23:11 - 23:13
the segment, this segment.
• 23:13 - 23:14
Are you guys with me?
• 23:14 - 23:17
Why don't I expand
to the whole line?
• 23:17 - 23:20
I say, I want to expand
to the whole line, which
• 23:20 - 23:21
would be stupid.
• 23:21 - 23:23
Why would it be stupid?
• 23:23 - 23:27
I would get y equals
something negative here.
• 23:27 - 23:31
And if I expand in the other
direction, x would be negative.
• 23:31 - 23:34
So it's not a good thing.
• 23:34 - 23:39
So the only thing I
have is the segment,
• 23:39 - 23:40
which has two endpoints.
• 23:40 - 23:42
Those two endpoints
are boo-boos.
• 23:42 - 23:45
• 23:45 - 23:47
The endpoints can give
you extrema as well.
• 23:47 - 23:49
We talked about it last time.
• 23:49 - 23:52
So every time you do
this, you're fine,
• 23:52 - 23:55
but you have to compare the
results against the extrema.
• 23:55 - 23:57
These are artificial cuts.
• 23:57 - 23:59
In what sense artificial?
• 23:59 - 24:04
In the sense that you
don't let the whole thing
• 24:04 - 24:07
evolve over the
whole real domain.
• 24:07 - 24:09
Once you artificially
cut something--
• 24:09 - 24:11
let me give you another example.
• 24:11 - 24:13
Don't put this in
the notes, because I
• 24:13 - 24:15
don't want it to confuse you.
• 24:15 - 24:18
You have some natural, so-called
relative max and minima here,
• 24:18 - 24:19
right?
• 24:19 - 24:22
That's a relative min, that's
a relative max, and so on.
• 24:22 - 24:27
If I make an artificial
cut anywhere-- let's
• 24:27 - 24:32
say this is not going
to be a minimum anymore.
• 24:32 - 24:36
I make an artificial cut here,
I make an artificial cut here.
• 24:36 - 24:39
These endpoints will
generate other possibilities
• 24:39 - 24:42
for my absolute max
and absolute min.
• 24:42 - 24:45
So those extrema are
extremely important.
• 24:45 - 24:47
I have one.
• 24:47 - 24:50
What is this guy's-- 1, 0.
• 24:50 - 24:51
Am I right?
• 24:51 - 24:54
And this is 0,1.
• 24:54 - 24:56
So I have to look
at the possibility.
• 24:56 - 24:59
When it's 1 by 1, it
goes-- say it again.
• 24:59 - 25:01
1,0.
• 25:01 - 25:03
And x2y2 was hmm?
• 25:03 - 25:05
0,1.
• 25:05 - 25:07
And of course, both
of them satisfy that.
• 25:07 - 25:13
In this case, f of 1, 0 has
to be evaluated as well.
• 25:13 - 25:18
That's going to be 1 minus
1 squared minus 0 equals 0.
• 25:18 - 25:22
And by the symmetry
of this polynomial,
• 25:22 - 25:29
you are going to have the
same answer, 0, in both cases.
• 25:29 - 25:31
You're going to draw the table.
• 25:31 - 25:35
And this is the perfect
place for the table.
• 25:35 - 25:38
Perfect place in the
sense that you have x, y
• 25:38 - 25:44
and you have-- who are your
notable, noticeable guys?
• 25:44 - 25:46
1/2, 1/2.
• 25:46 - 25:49
Who said 1,0?
• 25:49 - 25:53
And 0, 1.
• 25:53 - 25:57
And who was the zz
was the 1/2 here.
• 25:57 - 25:59
And here was 0, and here was 0.
• 25:59 - 26:03
And I'm going to start
making faces and drawing.
• 26:03 - 26:07
• 26:07 - 26:08
• 26:08 - 26:11
Did I solve this
problem at home?
• 26:11 - 26:13
Yes, I did.
• 26:13 - 26:14
And I got the same answer.
• 26:14 - 26:15
All right.
• 26:15 - 26:19
So this is max.
• 26:19 - 26:21
This is min.
• 26:21 - 26:23
This is mean, the same.
• 26:23 - 26:26
So both of these are what?
• 26:26 - 26:28
Absolute minima.
• 26:28 - 26:32
And these are the
absolute extrema
• 26:32 - 26:39
for this problem
with constraint.
• 26:39 - 26:43
and erase and say,
• 26:43 - 26:46
remember in the
• 26:46 - 26:49
how much work it was to do this?
• 26:49 - 26:52
And I'm going to apply
the other method.
• 26:52 - 26:54
So how much space?
• 26:54 - 26:59
So we needed one
board largely written.
• 26:59 - 27:02
You want to go to follow
the steps one and two.
• 27:02 - 27:03
Should I erase that?
• 27:03 - 27:03
STUDENT: No.
• 27:03 - 27:06
• 27:06 - 27:08
PROFESSOR: You
are my note-taker.
• 27:08 - 27:10
Of course I will listen to you.
• 27:10 - 27:16
And then let's see what method
number two I had in mind
• 27:16 - 27:18
is the one from last time.
• 27:18 - 27:19
So this is a what?
• 27:19 - 27:24
It's a review of-- what
is the section time?
• 27:24 - 27:26
11.7.
• 27:26 - 27:31
And I'm going to make a face,
happy that I can have yet
• 27:31 - 27:35
another application for you.
• 27:35 - 27:39
When this problem appeared on
the final at least five times
• 27:39 - 27:46
in the last 10 finals or
more, different instructors
• 27:46 - 27:49
viewed it differently.
• 27:49 - 27:53
Practically, the
general instruction
• 27:53 - 27:55
given to the students--
solve it anyway you
• 27:55 - 27:57
find it easier for you.
• 27:57 - 27:59
Just don't make mistakes.
• 27:59 - 28:03
So we did not
encourage instructors
• 28:03 - 28:05
to say, do this by
Lagrange multipliers,
• 28:05 - 28:09
or do this by-- no, no, no, no.
• 28:09 - 28:10
Whatever is easier
for the student.
• 28:10 - 28:13
So what did we do last
• 28:13 - 28:17
Since x and y are
married, y depend on x.
• 28:17 - 28:20
So y is 1 times x.
• 28:20 - 28:22
And we say, this
is my guy that I
• 28:22 - 28:26
have to plug in
into the function,
• 28:26 - 28:28
into the original function.
• 28:28 - 28:30
And then f would
not be a function
• 28:30 - 28:32
of two independent
variables anymore.
• 28:32 - 28:36
But it's going to become a
function of one variable.
• 28:36 - 28:38
Thank god it's not hard.
• 28:38 - 28:41
It's no hard
because in this way,
• 28:41 - 28:47
you have just to pull
out the y1 minus x,
• 28:47 - 28:51
and square it, and
do the algebra.
• 28:51 - 28:54
So 1 minus x squared.
• 28:54 - 28:56
And I'm going to do
this really quickly.
• 28:56 - 28:59
Minus 1 minus x
squared and plus 2x.
• 28:59 - 29:04
• 29:04 - 29:04
And OK.
• 29:04 - 29:10
So we say, all right, all
right, so 1 and minus 1 go away.
• 29:10 - 29:14
• 29:14 - 29:19
It make our life easier,
because I have minus 2x squared
• 29:19 - 29:21
plus So of course,
I could do it fast,
• 29:21 - 29:24
but the whole idea
is not to amaze you
• 29:24 - 29:30
with my capability of working
fast, but be able to follow.
• 29:30 - 29:32
So you have minus what?
• 29:32 - 29:33
So tell me.
• 29:33 - 29:36
You can pull out a minus 2x.
• 29:36 - 29:40
And you get x.
• 29:40 - 29:41
And a minus 1.
• 29:41 - 29:44
• 29:44 - 29:51
And what is special about that?
• 29:51 - 29:54
Well, do I really
need to do that?
• 29:54 - 29:56
That's the question.
• 29:56 - 29:57
Could I have stopped here?
• 29:57 - 30:00
Is this the point
of factoring out?
• 30:00 - 30:03
Not really because factoring
• 30:03 - 30:07
What I want is to chase
after Mr. f prime of x
• 30:07 - 30:13
and solve the critical point
equation f prime of x equals 0.
• 30:13 - 30:15
Right?
• 30:15 - 30:22
I need to find that x0 that will
satisfy f prime of x equals 0.
• 30:22 - 30:22
What do I get?
• 30:22 - 30:30
I get minus 4x plus 2 equals 0.
• 30:30 - 30:33
And I see I'm already relieved.
• 30:33 - 30:38
The moment I saw that, I
felt that I'm doing this
• 30:38 - 30:43
the right way, because I had
the previous method that led me
• 30:43 - 30:48
to a 1/2 that Alex provided for
somebody for the first time.
• 30:48 - 30:51
So now I feel I'm going
to get the same thing.
• 30:51 - 30:53
Let's see how much faster
or how much slower.
• 30:53 - 30:57
Why 0 corresponding
to it will be 1/2?
• 30:57 - 31:00
Because 1/2 plus 1/2 is a 1.
• 31:00 - 31:02
So what do I do?
• 31:02 - 31:05
Just as before, I start
my table and I say,
• 31:05 - 31:10
x and y must be 1/2 and 1/2
to give me the critical point
• 31:10 - 31:11
in the middle.
• 31:11 - 31:14
And I'm going to
get a 1/2 for that.
• 31:14 - 31:15
And I don't yet.
• 31:15 - 31:16
I pretend.
• 31:16 - 31:19
I don't know that's
gonna be a maximum.
• 31:19 - 31:25
What other points will provide
the books, the so-called--
• 31:25 - 31:27
STUDENT: Endpoints.
• 31:27 - 31:28
PROFESSOR: The endpoints.
• 31:28 - 31:31
And for those endpoints,
I keep in mind
• 31:31 - 31:34
that x and y,
again, are positive.
• 31:34 - 31:38
I should keep this picture in
mind, because if I don't, well,
• 31:38 - 31:40
it's not going to be very good.
• 31:40 - 31:44
So x is not allowed to move.
• 31:44 - 31:50
See, x has limited freedom
from the constraint.
• 31:50 - 31:54
So he's not allowed
to leave this segment.
• 31:54 - 31:56
x is going to be
between 0 and 1.
• 31:56 - 32:03
So for the endpoint x equals
0-- will provide me with y
• 32:03 - 32:04
equals 1.
• 32:04 - 32:07
And I'll put it in the
table, and I'll say,
• 32:07 - 32:11
when x equals 0 and y
equals 1-- and in that case,
• 32:11 - 32:15
I plug back in here and I get 0.
• 32:15 - 32:18
And again, for the
same type of-- I mean,
• 32:18 - 32:24
the other endpoint, I
get 1,0, and I get 0.
• 32:24 - 32:26
And it's the same thing.
• 32:26 - 32:29
I got the same thing
through another method.
• 32:29 - 32:33
This is the max, and
these are the mins.
• 32:33 - 32:39
And one of my students asked me
in my office hour-- by the way,
• 32:39 - 32:46
if you cannot make it to
Tuesday's 3:00 to 5:00,
• 32:46 - 32:47
you can come today.
• 32:47 - 32:50
At 2:00 after we are done,
I'm going to be in my office
• 32:50 - 32:51
as well.
• 32:51 - 32:53
So just.
• 32:53 - 32:58
So I have Tuesdays and Thursdays
after class, right after class.
• 32:58 - 33:04
Now, no matter what, if you get
the same answer, what if you
• 33:04 - 33:07
forget about one of the values?
• 33:07 - 33:11
what if I got the right maximum
• 33:11 - 33:15
and I got the right minimum, and
I say those are your extrema,
• 33:15 - 33:19
and I don't prove, mind you,
both points when it happens,
• 33:19 - 33:22
only one?
• 33:22 - 33:24
I don't know.
• 33:24 - 33:25
It's different from a
problem to the other.
• 33:25 - 33:28
Maybe I'm subtracting
some credit.
• 33:28 - 33:32
But you get most of the
partial credit in that case.
• 33:32 - 33:34
There will be many
values in which
• 33:34 - 33:37
you get the same altitude.
• 33:37 - 33:38
This is the altitude.
• 33:38 - 33:39
My z.
• 33:39 - 33:42
• 33:42 - 33:43
Do you have questions of that?
• 33:43 - 33:44
OK.
• 33:44 - 33:48
Now it's my turn
to make you vote.
• 33:48 - 33:51
And if you cannot
vote, you abstain.
• 33:51 - 33:53
Which one was easier?
• 33:53 - 33:57
The first method, the
Lagrange multipliers?
• 33:57 - 34:02
Or the second one, the-- how
should I call the second one,
• 34:02 - 34:03
the--
• 34:03 - 34:04
STUDENT: Integration.
• 34:04 - 34:08
PROFESSOR: The ray substitution
method then derivation,
• 34:08 - 34:11
count one type method?
• 34:11 - 34:15
So who is for-- OK.
• 34:15 - 34:19
You got this on the
midterm, say, or final.
• 34:19 - 34:22
How many of you would feel the
first method would be easier
• 34:22 - 34:24
to employ?
• 34:24 - 34:27
STUDENT: The second.
• 34:27 - 34:29
PROFESSOR: And how many of
you think the second method
• 34:29 - 34:32
is easier to employ?
• 34:32 - 34:35
And how many people
say that they
• 34:35 - 34:39
are equally long, or short,
or how many people abstain?
• 34:39 - 34:41
STUDENT: I would say it
depends on the problem.
• 34:41 - 34:43
PROFESSOR: Yeah, absolutely.
• 34:43 - 34:45
this particular one,
• 34:45 - 34:45
because I'm curious.
• 34:45 - 34:46
STUDENT: Oh, on this one.
• 34:46 - 34:48
Oh, OK.
• 34:48 - 34:50
PROFESSOR: I'm going to go
on and do another problem.
• 34:50 - 34:52
And for that, also, I will ask.
• 34:52 - 34:57
• 34:57 - 35:02
with other problems, it
may be that it's easier
• 35:02 - 35:05
to solve the system for
the Lagrange multipliers
• 35:05 - 35:11
than it is to pull out the y
explicitly from the constraint
• 35:11 - 35:13
and put it back.
• 35:13 - 35:19
• 35:19 - 35:22
What else have I
prepared for you?
• 35:22 - 35:24
• 35:24 - 35:27
• 35:27 - 35:30
some extra credit.
• 35:30 - 35:32
But I don't know
if you have time.
• 35:32 - 35:33
But write it anyway.
• 35:33 - 35:35
down, for one point
• 35:35 - 35:45
extra credit for
the next seven days,
• 35:45 - 35:54
of the following methods--
• 35:54 - 36:05
Lagrange multipliers with
one parameter lambda,
• 36:05 - 36:08
which is exactly the
same I taught you.
• 36:08 - 36:12
Same I taught.
• 36:12 - 36:17
And one that is not required
for the examinations, which
• 36:17 - 36:20
is Lagrange multipliers
with two parameters.
• 36:20 - 36:24
• 36:24 - 36:26
And that is a big
• 36:26 - 36:30
do that, because you
have two parameters.
• 36:30 - 36:33
Let's call them lambda and mu.
• 36:33 - 36:36
I don't know what to call them.
• 36:36 - 36:40
When you have that kind
of method, it's longer.
• 36:40 - 36:45
So it may take you several
pages of computation
• 36:45 - 36:49
to get to the lambdas and to
the extrema and everything.
• 36:49 - 36:52
But I would like you to
• 36:52 - 36:56
and write down a short
• 36:56 - 36:59
So both of them are one point.
• 36:59 - 37:03
Both of them are one point
extra credit at the end.
• 37:03 - 37:04
STUDENT: Together or each?
• 37:04 - 37:05
PROFESSOR: Yes, sir?
• 37:05 - 37:06
STUDENT: One point each
or one point together?
• 37:06 - 37:08
PROFESSOR: No, one
both, together.
• 37:08 - 37:08
I'm sorry.
• 37:08 - 37:12
Because there will be other
chances to get extra credit.
• 37:12 - 37:16
And I'm cooking up something
I didn't say on the syllabus,
• 37:16 - 37:21
like a brownie point
or cake or something.
• 37:21 - 37:24
At the end of the
class, I would like
• 37:24 - 37:28
you to write me a statement
of two pages on how
• 37:28 - 37:32
you think Calculus 3
• 37:32 - 37:35
And one question from
a previous student
• 37:35 - 37:39
was, I've changed
my major four times.
• 37:39 - 37:41
Which one shall I pick?
• 37:41 - 37:44
I said, whichever
you are in right now.
• 37:44 - 37:45
How does that relate?
• 37:45 - 37:49
How is Calculus 3
• 37:49 - 37:51
Give me some
examples and how you
• 37:51 - 37:55
think functions of two
variables or three variables--
• 37:55 - 37:56
STUDENT: What's this?
• 37:56 - 38:00
PROFESSOR: Up here
• 38:00 - 38:03
STUDENT: So it's a two
parameter question?
• 38:03 - 38:06
Like, would there be any
question regarding that?
• 38:06 - 38:07
PROFESSOR: No, nothing.
• 38:07 - 38:08
Not in the homework.
• 38:08 - 38:12
We don't cover that, we
don't do that in the test.
• 38:12 - 38:14
Most instructors
don't even mention it,
• 38:14 - 38:17
but I said, mm, you
are our students,
• 38:17 - 38:20
so I want to let you do
a little bit of research.
• 38:20 - 38:23
• 38:23 - 38:24
Individual study.
• 38:24 - 38:26
STUDENT: Is that in the book?
• 38:26 - 38:27
PROFESSOR: It is in the book.
• 38:27 - 38:28
So individual study.
• 38:28 - 38:30
One page or one page and a half.
• 38:30 - 38:31
Something like that.
• 38:31 - 38:33
Maybe less.
• 38:33 - 38:33
OK.
• 38:33 - 38:36
• 38:36 - 38:43
One other one that I cooked
up-- it's not in the book.
• 38:43 - 38:47
But I liked it because it sounds
like a real-life application.
• 38:47 - 38:50
It is a real-life application.
• 38:50 - 38:55
And I was talking
to the mailman.
• 38:55 - 39:00
And he was saying, I wonder
how-- because a guy, poor guy,
• 39:00 - 39:02
was carrying these
Priority boxes.
• 39:02 - 39:06
And he said, I wonder
how they optimize?
• 39:06 - 39:09
When they say "flat
rate," how do they
• 39:09 - 39:12
come up with those dimensions?
• 39:12 - 39:15
And it's an
optimization problem,
• 39:15 - 39:19
and there are many like
that in the real world.
• 39:19 - 39:23
But for my case,
I would say, let's
• 39:23 - 39:27
assume that somebody says,
the sum of the lengths
• 39:27 - 39:33
plus widths plus height is
constrained to be some number.
• 39:33 - 39:37
x plus y plus z equals
the maximum possible.
• 39:37 - 39:39
Could be 50 inches.
• 39:39 - 39:41
inches-- because I
• 39:41 - 39:43
don't want to work with
that kind of numbers,
• 39:43 - 39:48
I'm too lazy-- I put x
plus y plus equals 1.
• 39:48 - 39:50
That's my constraint, g.
• 39:50 - 39:58
• 39:58 - 40:03
I would like to
maximize the volume.
• 40:03 - 40:04
Say it again, Magdalena.
• 40:04 - 40:05
What is the problem?
• 40:05 - 40:07
• 40:07 - 40:11
My problem is example three.
• 40:11 - 40:34
Maximize the volume of a box
of length, height, and width x,
• 40:34 - 40:41
y, z, just to make our
life easier in a way
• 40:41 - 40:46
that the girth cross
the-- well, OK.
• 40:46 - 40:49
Let me make this interesting.
• 40:49 - 40:54
The sum of the
dimensions equals 1.
• 40:54 - 40:56
And where can you
find this problem?
• 40:56 - 41:00
Well, this problem can be
found in several resources.
• 41:00 - 41:03
We haven't dealt very
much with functions
• 41:03 - 41:06
of three variables, x, y, z.
• 41:06 - 41:09
But the procedure
is exactly the same.
• 41:09 - 41:12
I stole that from
a library online
• 41:12 - 41:17
that's called Paul's
Online Calculus Notes.
• 41:17 - 41:20
And imagine that
the same thing I
• 41:20 - 41:24
taught you would be applied to
functions of three variables.
• 41:24 - 41:27
Tell me who the volume will be.
• 41:27 - 41:30
The volume would be a
function of three variables.
• 41:30 - 41:33
Let's call it f, which is what?
• 41:33 - 41:35
Who is telling me what?
• 41:35 - 41:36
STUDENT: x times y.
• 41:36 - 41:37
PROFESSOR: x times y.
• 41:37 - 41:38
Thanks.
• 41:38 - 41:40
And are we happy about it?
• 41:40 - 41:42
Ah, it's a beautiful function.
• 41:42 - 41:44
It's not going to give you
• 41:44 - 41:47
• 41:47 - 41:53
I would like you to cook up
step one and step two for me
• 41:53 - 41:56
by the Lagrange
multipliers I specify.
• 41:56 - 42:05
• 42:05 - 42:07
For functions of
three variables.
• 42:07 - 42:14
• 42:14 - 42:15
Maximize and minimize.
• 42:15 - 42:20
• 42:20 - 42:20
Yeah.
• 42:20 - 42:25
• 42:25 - 42:25
OK.
• 42:25 - 42:32
going to be in R2 anymore.
• 42:32 - 42:33
They will be in R3.
• 42:33 - 42:34
And so what?
• 42:34 - 42:35
It doesn't matter.
• 42:35 - 42:35
Step one.
• 42:35 - 42:40
• 42:40 - 42:41
Say it again, Magdalena.
• 42:41 - 42:41
What do you mean?
• 42:41 - 42:45
I mean that when you're going
to have something like that,
• 42:45 - 42:52
the system for nabla f of
x, y, z at the point x0, y0,
• 42:52 - 43:01
z0 will be lambda times
nabla g of x at x0, y0 is 0,
• 43:01 - 43:05
where both nablas are in R3.
• 43:05 - 43:05
Right?
• 43:05 - 43:12
They will be f sub x, f sub
y, f sub z angular brackets.
• 43:12 - 43:17
two equations in the system,
• 43:17 - 43:18
you're going to have
three equations.
• 43:18 - 43:21
That's the only big difference.
• 43:21 - 43:22
Big deal.
• 43:22 - 43:24
Not a big deal.
• 43:24 - 43:26
So you tell me what
I'm going to write.
• 43:26 - 43:34
So I'm going to write f sub
x equals lambda g sub x.
• 43:34 - 43:37
f sub y equals lambda g sub y.
• 43:37 - 43:41
f sub z equals lambda g sub z.
• 43:41 - 43:45
Thank god I don't have
more than three variables.
• 43:45 - 43:49
Now we-- in fact, it's
how do you think engineers
• 43:49 - 43:51
solve this kind of system?
• 43:51 - 43:52
Do they do this by hand?
• 43:52 - 43:53
No.
• 43:53 - 43:55
Life is complicated.
• 43:55 - 43:59
When you do Lagrange multipliers
on a thermodynamical problem
• 43:59 - 44:02
or mechanics problem,
physics problem,
• 44:02 - 44:06
you have really ugly
data that are programs
• 44:06 - 44:08
based on Lagrange multipliers.
• 44:08 - 44:10
You can have a
Lagrange multiplier
• 44:10 - 44:13
of seven different parameters,
including pressure, time,
• 44:13 - 44:14
and temperature.
• 44:14 - 44:16
And it's really horrible.
• 44:16 - 44:19
And you don't do that by hand.
• 44:19 - 44:23
That's why we have to be
thankful to technology
• 44:23 - 44:27
and the software, the
scientific software methods.
• 44:27 - 44:30
You can do that in MATLAB, you
can do that in Mathematica.
• 44:30 - 44:32
MATLAB is mostly for engineers.
• 44:32 - 44:35
There are programs written
especially for MATLAB
• 44:35 - 44:40
to solve the problem of
Lagrange multipliers.
• 44:40 - 44:43
Now, this has not complete.
• 44:43 - 44:47
We are missing the most
important, the marriage thing,
• 44:47 - 44:53
the g of x, y, z constraint.
• 44:53 - 44:55
Now there are three
in the picture.
• 44:55 - 45:01
I don't know what that means.
x plus y plus z equals 1.
• 45:01 - 45:05
• 45:05 - 45:09
So if and only if, who's going
to tell me what those will be?
• 45:09 - 45:10
Are they going to be hard?
• 45:10 - 45:12
No.
• 45:12 - 45:16
It's a real-life problem, but
it's not a hard problem. f of x
• 45:16 - 45:19
will be yz equals lambda.
• 45:19 - 45:20
Who is g sub x?
• 45:20 - 45:21
STUDENT: 1.
• 45:21 - 45:22
PROFESSOR: 1.
• 45:22 - 45:23
Thank god.
• 45:23 - 45:24
So it's fine.
• 45:24 - 45:26
It's not that.
• 45:26 - 45:30
F sub y would be
xz equals lambda.
• 45:30 - 45:34
f sub z is xy equals lambda.
• 45:34 - 45:39
Ah, there is a lot
of symmetry in that.
• 45:39 - 45:41
I have some thinking to do.
• 45:41 - 45:43
Well, I'm a scientist.
• 45:43 - 45:47
I have to take into account
all the possibilities.
• 45:47 - 45:50
If I lose one, I'm dead
meat, because that one
• 45:50 - 45:52
may be essential.
• 45:52 - 45:56
So if I were a computer,
I would branch out
• 45:56 - 45:59
all the possibilities
in a certain order.
• 45:59 - 46:00
But I'm not a computer.
• 46:00 - 46:04
But I have to think in
the same organized way
• 46:04 - 46:09
to exhaust all the
possibilities for that.
• 46:09 - 46:13
And for that matter, I
have to pay attention.
• 46:13 - 46:17
So I have x plus
y plus z equals 1.
• 46:17 - 46:19
OK.
• 46:19 - 46:22
I'm going to give you
• 46:22 - 46:23
Yes.
• 46:23 - 46:24
I'll give you two
minutes to think
• 46:24 - 46:28
how to solve-- how
does one solve that?
• 46:28 - 46:33
How does one solve it?
• 46:33 - 46:35
Think how you would grab.
• 46:35 - 46:37
Where would you grab
the problem from?
• 46:37 - 46:40
But think it for yourself, and
then I'm gone for two minutes.
• 46:40 - 46:43
And then I'm going to
discuss things out loud,
• 46:43 - 46:47
and I'll share with
you how I did it.
• 46:47 - 46:51
STUDENT: It could
be 1, 1 minus 1.
• 46:51 - 46:53
PROFESSOR: You are
like an engineer.
• 46:53 - 47:00
I could have some equality
• 47:00 - 47:02
between the coordinate.
• 47:02 - 47:05
We have to do it in
a mathematical way.
• 47:05 - 47:06
All right?
• 47:06 - 47:12
So would it help me if I
subtracted the second equation
• 47:12 - 47:13
from the first equation?
• 47:13 - 47:15
What kind of
information would I get?
• 47:15 - 47:17
STUDENT: But that
• 47:17 - 47:18
STUDENT: We can divide better.
• 47:18 - 47:20
• 47:20 - 47:21
PROFESSOR: I can divide.
• 47:21 - 47:23
That's another possibility.
• 47:23 - 47:28
I can divide and
do y/x equals 1.
• 47:28 - 47:33
And that would give
you x equals y.
• 47:33 - 47:34
And then you plug it back.
• 47:34 - 47:36
And then you say, wait a minute.
• 47:36 - 47:40
If x equals y, then
x times x is lambda.
• 47:40 - 47:44
So lambda would be x squared.
• 47:44 - 47:46
So then we plug it in here.
• 47:46 - 47:50
And we go, x plus x equals 2x.
• 47:50 - 47:54
And then we see what else we
can find that information.
• 47:54 - 47:58
As you can see, there is no
unique way of doing that.
• 47:58 - 48:00
But what's unique
• 48:00 - 48:06
No matter how I do it, I should
overlap with Nitish's method.
• 48:06 - 48:09
At some point, I should
get the possibility
• 48:09 - 48:10
that x and y are the same.
• 48:10 - 48:13
If I don't, that means
I'm doing something wrong.
• 48:13 - 48:18
So the way I approach this
problem-- OK, one observation,
• 48:18 - 48:21
I could subtract the second
from the first, where
• 48:21 - 48:23
I would subtract the
third from the second.
• 48:23 - 48:27
Or I could subtract the
second from the first
• 48:27 - 48:29
and analyze all
the possibilities.
• 48:29 - 48:33
Let's do only one
and then by symmetry,
• 48:33 - 48:35
because this is a
symmetric problem.
• 48:35 - 48:37
By symmetry, I'm going to
see all the other problems.
• 48:37 - 48:41
So how do you think in symmetry?
• 48:41 - 48:46
x and y and z have-- it's a
democratic world for them.
• 48:46 - 48:48
They have the same roles.
• 48:48 - 48:51
So at some point when you
got some solutions for x, y,
• 48:51 - 48:55
z in a certain way,
you may swap them.
• 48:55 - 48:58
You may change the
rules of x, y, z,
• 48:58 - 49:00
and get all the solutions.
• 49:00 - 49:08
So the way I did it was I took
first xz minus yz equals 0.
• 49:08 - 49:11
• 49:11 - 49:16
But then let's interpret what
this-- and a mathematician
• 49:16 - 49:20
will go either by if and
only-- if/or it implies.
• 49:20 - 49:23
I don't know if
anybody taught you.
• 49:23 - 49:26
Depends where
you're coming from,
• 49:26 - 49:28
because different
schools, different states,
• 49:28 - 49:30
different customs
for this differently.
• 49:30 - 49:35
But in professional mathematics,
one should go with if and only
• 49:35 - 49:41
if, or implication,
x minus yz equals 0.
• 49:41 - 49:44
• 49:44 - 49:46
And then what
implication do I have?
• 49:46 - 49:48
Now I don't have an implication.
• 49:48 - 49:54
I have it in the sense
that I have either/or.
• 49:54 - 49:59
So this will go, like in
computer science, either/or.
• 49:59 - 50:10
Either-- I do the branching--
x equals y, or z equals 0.
• 50:10 - 50:15
And I have to study
these cases separately.
• 50:15 - 50:17
You see?
• 50:17 - 50:19
It's not so obvious.
• 50:19 - 50:22
Let me take this one, because
it's closer in my area
• 50:22 - 50:23
on [INAUDIBLE].
• 50:23 - 50:26
It doesn't matter in
which order I start.
• 50:26 - 50:30
For z equals 0, if I plug in
z equals 0, what do I get?
• 50:30 - 50:32
Lambda equals 0, right?
• 50:32 - 50:35
• 50:35 - 50:38
But if lambda equals 0, I
get another ramification.
• 50:38 - 50:41
So you are going to say,
• 50:41 - 50:42
Not yet.
• 50:42 - 50:49
So lambda equals 0 will again
• 50:49 - 50:53
Either x equals 0 or--
• 50:53 - 50:54
STUDENT: y equals 0.
• 50:54 - 50:55
PROFESSOR: --y equals 0.
• 50:55 - 50:58
• 50:58 - 51:01
Let's take the first one.
• 51:01 - 51:04
Like a computer,
just like a computer,
• 51:04 - 51:07
computer will say,
if-- so I'm here.
• 51:07 - 51:12
If 0 is 0 and x was
0, what would y be?
• 51:12 - 51:14
y will be 1.
• 51:14 - 51:16
That is the only case I got.
• 51:16 - 51:20
And I make a smile, because
why do I make a smile now?
• 51:20 - 51:25
Because I got all three of
them, and I can start my table
• 51:25 - 51:27
that's a pink table.
• 51:27 - 51:34
And here I have x, y,
z significant values.
• 51:34 - 51:37
Everything else doesn't matter.
• 51:37 - 51:42
And this is z, which was the
volume, which was x, y, z.
• 51:42 - 51:43
Was it, guys?
• 51:43 - 51:48
So I have to compare volumes
for this thinking box.
• 51:48 - 51:49
Right?
• 51:49 - 51:50
OK.
• 51:50 - 51:55
Now, in this case, I have 0, x.
• 51:55 - 51:57
y is 1.
• 51:57 - 51:58
z is 0.
• 51:58 - 52:03
The volume will be--
and do I have a box?
• 52:03 - 52:05
No, I don't have a box.
• 52:05 - 52:07
I make a face like that.
• 52:07 - 52:09
But the value is
still there to put.
• 52:09 - 52:12
As a mathematician, I
have to record everything.
• 52:12 - 52:14
STUDENT: Do you have to
put this on the exam?
• 52:14 - 52:16
Because it doesn't make sense.
• 52:16 - 52:17
This would not--
• 52:17 - 52:17
PROFESSOR: No.
• 52:17 - 52:18
No.
• 52:18 - 52:23
Because I haven't said,
if the box cannot be used.
• 52:23 - 52:25
I didn't say I
would use it or not.
• 52:25 - 52:29
So the volume 0 is a possible
value for the function.
• 52:29 - 52:31
And that will give
us the minimum.
• 52:31 - 52:34
So what do we-- I expect
you to say in the exam,
• 52:34 - 52:37
I have the absolute minimum.
• 52:37 - 52:40
One of the points-- I'm
going to have more points
• 52:40 - 52:43
when I have minima.
• 52:43 - 52:43
OK.
• 52:43 - 52:44
And the other case.
• 52:44 - 52:46
I don't want to get
distracted. y is 0.
• 52:46 - 52:49
So I get x equals 1.
• 52:49 - 52:51
Are you guys with me?
• 52:51 - 52:53
From here and here
and here, I get
• 52:53 - 52:57
x equals 1, because the sum
of all three of them will be,
• 52:57 - 52:58
again, 1.
• 52:58 - 52:59
So I have another pair.
• 52:59 - 53:01
0, 0.
• 53:01 - 53:03
STUDENT: Wouldn't it be 1, 0?
• 53:03 - 53:06
PROFESSOR: 1, 0.
• 53:06 - 53:09
1, 0, 0.
• 53:09 - 53:12
And the volume will be the same.
• 53:12 - 53:15
And another absolute minima.
• 53:15 - 53:19
Remember that everything
is positive-- the x, y, z,
• 53:19 - 53:20
and the [INAUDIBLE].
• 53:20 - 53:23
I keep going.
• 53:23 - 53:28
And I say, how do I get-- I
have the feeling I'm going
• 53:28 - 53:30
to get 0, 0, 1 at some point.
• 53:30 - 53:33
But how am I going
to get this thing?
• 53:33 - 53:34
I'm going to get
to it naturally.
• 53:34 - 53:36
So I should never anticipate.
• 53:36 - 53:39
• 53:39 - 53:42
The other case
will give it to me.
• 53:42 - 53:43
OK?
• 53:43 - 53:44
So let's see.
• 53:44 - 53:48
When x equals y, I
didn't say anything.
• 53:48 - 53:52
When x equals y, I have
to see what happens.
• 53:52 - 53:57
And I got here again two cases.
• 53:57 - 54:02
Either x equals-- either,
Magdalena, either x
• 54:02 - 54:09
equals y equals 0, or x
equals y equals non-zero.
• 54:09 - 54:10
So I'm a robot.
• 54:10 - 54:12
I'm an android.
• 54:12 - 54:18
I don't let any logical
piece escape me.
• 54:18 - 54:22
Everything goes in
the right place.
• 54:22 - 54:25
When x equals y equals
0, the only possibility I
• 54:25 - 54:29
have is z to the 1.
• 54:29 - 54:30
And I make another face.
• 54:30 - 54:31
I'm why?
• 54:31 - 54:32
Happy that I'm at the end.
• 54:32 - 54:35
But then I realize
that it is, of course,
• 54:35 - 54:38
not what I hoped for.
• 54:38 - 54:40
It's another minimum.
• 54:40 - 54:43
So I have minima
0 for the volume
• 54:43 - 54:48
attained at all these three
possibilities, all the three
• 54:48 - 54:48
points.
• 54:48 - 54:53
• 54:53 - 54:53
And then what?
• 54:53 - 54:56
Then finally something
more interesting.
• 54:56 - 54:58
Finally.
• 54:58 - 55:00
x equals y different from 0.
• 55:00 - 55:02
What am I doing to
do with that case?
• 55:02 - 55:05
Of course, you can
do this in many ways.
• 55:05 - 55:14
But if you want to know what I
did, just don't laugh too hard.
• 55:14 - 55:16
I said, look, I'm
changing everything
• 55:16 - 55:18
in the original thing.
• 55:18 - 55:23
I'll take it aside, and
I'll plug in and see
• 55:23 - 55:25
what the system becomes.
• 55:25 - 55:30
So we'll assume x equals
y different from 0,
• 55:30 - 55:31
and plug it back in the system.
• 55:31 - 55:35
In that case, xy equals
lambda will become x squared
• 55:35 - 55:38
equals lambda, right?
• 55:38 - 55:41
Mr. x plus y plus
z equals 1 would
• 55:41 - 55:46
become x plus x plus z, which
is 2x plus z, which is 1.
• 55:46 - 55:49
• 55:49 - 55:53
And finally, these
two equations,
• 55:53 - 55:57
since x equals y are one and
the same, they become one.
• 55:57 - 55:59
xz equals lambda.
• 55:59 - 56:01
And I stare at this guy.
• 56:01 - 56:05
And somebody tell
me, can I solve that?
• 56:05 - 56:07
Well, it's a system,
not a linear system.
• 56:07 - 56:11
But it's a system
of three variables.
• 56:11 - 56:15
Three equations-- I'm sorry--
with three unknowns-- x, z,
• 56:15 - 56:15
and lambda.
• 56:15 - 56:20
So it should be easy
for me to solve it.
• 56:20 - 56:22
How did I solve it?
• 56:22 - 56:26
I got-- it's a little bit funny.
• 56:26 - 56:29
I got x equals lambda over z.
• 56:29 - 56:33
And then I went-- but let
me square the whole thing.
• 56:33 - 56:36
And I'm going to get--
why do I square it?
• 56:36 - 56:39
Because I want to compare
it to what I have here.
• 56:39 - 56:42
If I compare, I go, if
and only if x squared
• 56:42 - 56:48
equals lambda squared
over z squared.
• 56:48 - 56:52
But Mr. x squared is
known as being lambda.
• 56:52 - 56:55
So I will replace
him. x squared is
• 56:55 - 56:56
lambda from the first equation.
• 56:56 - 57:03
So I get lambda equals lambda
squared over z squared.
• 57:03 - 57:07
So I got that-- what did I get?
• 57:07 - 57:08
Nitish, tell me.
• 57:08 - 57:10
Lambda equals?
• 57:10 - 57:11
STUDENT: x squared.
• 57:11 - 57:11
PROFESSOR: z squared.
• 57:11 - 57:13
STUDENT: Lambda is
equal to z squared.
• 57:13 - 57:19
PROFESSOR: So if and only
if lambda equals z squared.
• 57:19 - 57:23
But lambda was x
squared as well.
• 57:23 - 57:24
So lambda was what?
• 57:24 - 57:29
Lambda was z squared,
and lambda was x squared.
• 57:29 - 57:34
And it implies that x equals z.
• 57:34 - 57:36
x is equal to z.
• 57:36 - 57:38
But it's also equal to y.
• 57:38 - 57:40
Alex jump on me.
• 57:40 - 57:41
Why would that be?
• 57:41 - 57:43
STUDENT: Because
you just said that--
• 57:43 - 57:45
PROFESSOR: Because x was
y from the assumption.
• 57:45 - 57:46
So equal to y.
• 57:46 - 57:50
So this is the beautiful thing,
where all the three dimensions
• 57:50 - 57:51
are the same.
• 57:51 - 57:55
• 57:55 - 57:59
So what do we know that
thingie-- x equals z equals y?
• 57:59 - 58:01
STUDENT: It's a box.
• 58:01 - 58:02
PROFESSOR: It's a box of a what?
• 58:02 - 58:03
STUDENT: It's a square.
• 58:03 - 58:03
STUDENT: Square.
• 58:03 - 58:03
PROFESSOR: It's a--
• 58:03 - 58:04
ALL STUDENTS: Cube.
• 58:04 - 58:05
PROFESSOR: Cube.
• 58:05 - 58:05
OK.
• 58:05 - 58:06
So for the cube--
• 58:06 - 58:07
STUDENT: Square box.
• 58:07 - 58:10
PROFESSOR: We get--
for z, they were
• 58:10 - 58:13
dimensions we can have.
• 58:13 - 58:17
So they said, x plus y plus
z should be, at most, 1.
• 58:17 - 58:23
But we managed to maximize
the volume by the cube.
• 58:23 - 58:27
The cube is the only one
that maximizes the volume.
• 58:27 - 58:29
How do I get it back?
• 58:29 - 58:34
So I get it back by saying,
x plus y plus z equals 1.
• 58:34 - 58:39
So the only possibility that
comes out from here is that--
• 58:39 - 58:40
STUDENT: They're all 1/3.
• 58:40 - 58:45
PROFESSOR: That I
have 1/3, 1/3, 1/3.
• 58:45 - 58:48
And I have to take
this significant point.
• 58:48 - 58:51
This is the significant
point that I was praying for.
• 58:51 - 58:55
And the volume will be 1/27.
• 58:55 - 58:58
• 58:58 - 58:59
And I'm happy.
• 58:59 - 59:01
Why am I so happy?
• 59:01 - 59:04
Is this 1/27 the best I can get?
• 59:04 - 59:07
In this case, yes.
• 59:07 - 59:10
So I have the maximum.
• 59:10 - 59:12
Now, assume that
somebody would have--
• 59:12 - 59:13
STUDENT: That's a
really small box.
• 59:13 - 59:14
PROFESSOR: It's a small box.
• 59:14 - 59:15
Exactly.
• 59:15 - 59:18
I'm switching to
something, so assume--
• 59:18 - 59:22
I don't know why airlines
do that, but they do.
• 59:22 - 59:27
They say, the girth plus
the height will be this.
• 59:27 - 59:31
Girth meaning-- the
girth would be--
• 59:31 - 59:35
so this is the height
of your-- can I get?
• 59:35 - 59:36
Or this one?
• 59:36 - 59:36
No, that's yours.
• 59:36 - 59:38
Oh.
• 59:38 - 59:39
It's heavy.
• 59:39 - 59:41
You shouldn't make
me carry this.
• 59:41 - 59:41
OK.
• 59:41 - 59:46
So x plus y plus x
plus y is the girth.
• 59:46 - 59:48
And some airlines
are really weird.
• 59:48 - 59:50
I've dealt with at least
12 different airlines.
• 59:50 - 59:55
And the low-cost airlines that
I've dealt with in Europe,
• 59:55 - 59:57
they don't tell you what.
• 59:57 - 60:00
They say, maximum, 10-kilo max.
• 60:00 - 60:01
• 60:01 - 60:06
And the girth plus the length
has to be a certain thing.
• 60:06 - 60:09
And others say just the--
some of the three dimensions
• 60:09 - 60:11
should be something like that.
• 60:11 - 60:14
Whatever they give you.
• 60:14 - 60:18
So I know you don't think
in centimeters usually.
• 60:18 - 60:20
But imagine that
somebody gives you
• 60:20 - 60:23
the sum of the three
• 60:23 - 60:27
would be 100.
• 60:27 - 60:30
That is horrible.
• 60:30 - 60:35
What would be the maximum
volume in that case?
• 60:35 - 60:37
STUDENT: It would all be
33 and 1/3 centimeters.
• 60:37 - 60:37
PROFESSOR: Huh?
• 60:37 - 60:39
STUDENT: It would all be
33 and 1/3 centimeters.
• 60:39 - 60:40
PROFESSOR: You mean?
• 60:40 - 60:45
STUDENT: They'll all be 33 and
1/3 centimeters, x, y, and z.
• 60:45 - 60:48
PROFESSOR: Not the sum. x
plus y plus z would be 100.
• 60:48 - 60:49
STUDENT: So each one of them?
• 60:49 - 60:52
PROFESSOR: And then you
have 33.33 whatever.
• 60:52 - 60:56
And then you cube that,
and you get the volume.
• 60:56 - 60:58
Now, would that be practical?
• 60:58 - 60:59
STUDENT: No.
• 60:59 - 61:00
PROFESSOR: Why not?
• 61:00 - 61:04
• 61:04 - 61:05
STUDENT: It doesn't fit.
• 61:05 - 61:08
PROFESSOR: It doesn't the
• 61:08 - 61:11
So we try to-- because
• 61:11 - 61:14
sort of flattened out, we
have the flattened ones.
• 61:14 - 61:17
But in any case,
it's a hassle just
• 61:17 - 61:21
having to deal with
this kind of constraint.
• 61:21 - 61:24
And when you come back
to the United States,
• 61:24 - 61:28
you really feel-- I don't know
if you have this experience.
• 61:28 - 61:30
The problem is not in
between continents.
• 61:30 - 61:33
• 61:33 - 61:36
You have plenty of-- you
can check in a baggage.
• 61:36 - 61:39
But if you don't, which I
don't, because I'm really weird.
• 61:39 - 61:42
I get a big carry-on,
and I can fit that.
• 61:42 - 61:44
And I'm very happy.
• 61:44 - 61:48
I have everything I need for
three weeks to one month.
• 61:48 - 61:53
But if you deal with low-cost
airlines, on that kind of 70
• 61:53 - 61:59
euro or something between London
and Milan, or Paris, or London
• 61:59 - 62:05
and Athens, or something,
and you pay that little, they
• 62:05 - 62:07
have all sorts of weird
constraints like this one.
• 62:07 - 62:11
x plus y plus z has to
be no more than that.
• 62:11 - 62:14
And the weight should be
no more than 20 pounds.
• 62:14 - 62:16
And I'll see how
you deal with that.
• 62:16 - 62:17
It's not easy.
• 62:17 - 62:20
American airlines all the time,
• 62:20 - 62:25
but compared to those airlines,
we are really spoiled.
• 62:25 - 62:27
In the ticket price,
we are paying,
• 62:27 - 62:33
let's say, \$300 from
here to Memphis,
• 62:33 - 62:39
we have a lot of goodies
includes that we may not always
• 62:39 - 62:40
appreciate.
• 62:40 - 62:43
I'm not working for
American Airlines.
• 62:43 - 62:47
Actually, I prefer
Southwest a lot
• 62:47 - 62:50
by the way they treat
us customers and so on.
• 62:50 - 62:53
But I'm saying,
think of restrictions
• 62:53 - 62:57
when it comes to
volume and weight,
• 62:57 - 63:01
because they represent something
in real-life applications.
• 63:01 - 63:01
Yes, sir.
• 63:01 - 63:04
STUDENT: I have a question
• 63:04 - 63:06
I found the 1/3 just
by finding the ratio--
• 63:06 - 63:09
the first and the second and
then the second and the third.
• 63:09 - 63:09
PROFESSOR: That's how Nitish--
• 63:09 - 63:10
STUDENT: Yeah, that's
how I did it as well.
• 63:10 - 63:11
PROFESSOR: You were
napping a little bit.
• 63:11 - 63:12
But yeah.
• 63:12 - 63:13
But then you woke up.
• 63:13 - 63:15
[LAUGHTER]
• 63:15 - 63:18
While you were napping, he goes,
divide by the first equation
• 63:18 - 63:20
by the second one,
and you get 1.
• 63:20 - 63:22
x/y is 1.
• 63:22 - 63:27
And so you get the solution
of having all of them equal,
• 63:27 - 63:27
all three.
• 63:27 - 63:28
STUDENT: Yeah.
• 63:28 - 63:29
Just because I did that
out, and then I was like,
• 63:29 - 63:32
oh, it's y is equal to x,
y is equal to z, and then
• 63:32 - 63:33
just change it all to y.
• 63:33 - 63:33
PROFESSOR: Right.
• 63:33 - 63:37
So how do you think I'm going
• 63:37 - 63:38
Do I care?
• 63:38 - 63:39
No.
• 63:39 - 63:42
As long as you get the right
• 63:42 - 63:45
I don't care which
method you were using.
• 63:45 - 63:51
The problem for me comes where
you have had the right idea.
• 63:51 - 63:54
You messed up in the
middle of the algebra,
• 63:54 - 63:58
and you gave me the
wrong algebraic solution.
• 63:58 - 64:01
That's where I have to ponder
how much partial credit
• 64:01 - 64:03
I want to give you
or not give you.
• 64:03 - 64:09
But I'm trying to be fair,
in those cases, to everybody.
• 64:09 - 64:12
I wanted to tell you-- I
don't know if you realize,
• 64:12 - 64:17
but I stole from you every
• 64:17 - 64:20
• 64:20 - 64:23
It should be a little
bit cumulative.
• 64:23 - 64:25
I'm going to give you
back those minutes
• 64:25 - 64:28
right now, hoping that
those seven minutes,
• 64:28 - 64:32
you're going to use them doing
something useful for yourself.
• 64:32 - 64:35
• 64:35 - 64:39
At the same time, I'm
• 64:39 - 64:43
either now, either here,
or in my office upstairs.
• 64:43 - 64:47
And I know many of you
solved a lot of the homework.
• 64:47 - 64:49
I'm proud of you.
• 64:49 - 64:50
Some of you did not.
• 64:50 - 64:52
Some of you still struggle.
• 64:52 - 64:54
• 64:54 - 64:58
Is it too early-- I
• 64:58 - 65:03
can I have the homework early?
• 65:03 - 65:04
Is it too early?
• 65:04 - 65:06
I don't know what to do.
• 65:06 - 65:08
I mean, I feel it's
too early to give you
• 65:08 - 65:11
Chapter 12 [INAUDIBLE]
and Chapter
• 65:11 - 65:14
• 65:14 - 65:21
But if you feel it's OK, I can
send you the homework next.
• 65:21 - 65:22
Yeah?
• 65:22 - 65:22
All right.
• 65:22 - 65:24
Whatever you want.
• 65:24 - 65:26
We will start
Chapter 12 next week.
• 65:26 - 65:30
• 65:30 - 65:35
and I can go ahead and start
the homework for Chapter 12
• 65:35 - 65:36
• 65:36 - 65:38
And keep it for a month or so.
• 65:38 - 65:43
I feel that as long as you
don't procrastinate, it's OK.
• 65:43 - 65:47
STUDENT: I solved that one,
because I've seen it before.
Title:
TTU Math2450 Calculus3 Sec 11.7 and 11.8
Description:

Absolute extrema and Lagrange multipliers

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Video Language:
English
 jackie.luft edited English subtitles for TTU Math2450 Calculus3 Sec 11.7 and 11.8