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JOEL LEWIS: Hi.

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Welcome back to recitation.

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One of the things you've been
learning about in lecture is

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how to solve some max-min
problems. How to find the

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maximum or minimum of a given
function given some

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constraints and figure out where
it reaches its largest

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or smallest value.

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So here I have a particular
example of such a problem.

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So we're going to build
a cardboard box.

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And our cardboard box has to
meet the following criteria.

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So its volume has
to be 3 units.

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The front and back of the
cardboard box are going to be

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made just of single-thickness
cardboard.

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But the two sides, the left
and right, are going to be

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made double-thick.

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And the bottom is going to
be made triple-thick.

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We're not going to have
a top, it's just going

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to be an open box.

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So it's going to have five
sides: two of single

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thickness, two of double
thickness, and with the bottom

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of triple thickness.

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So, the question is-- there
are, you know, a lot of

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different shapes of box with
total volume 3 units.

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So the question is what
dimensions should this box

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have so that we use
the minimum amount

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of cardboard possible?

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All right?

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So we want the dimensions of
the box that use the least

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total amount cardboard.

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So why don't you pause the
video, work on that for

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awhile, come back, and we
can work on it together

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I hope you had some fun
puzzling this one out.

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Let's have a go at it.

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So to start, I think I'd just
like to draw a little

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picture of a box.

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So here's my cardboard box.

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Its top is going to
be open, say.

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So let's give it--

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we want to figure out what its
dimensions should be in order

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to minimize some expression,
so let's give them names.

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Right?

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So I think a natural thing to
call them is we can call one

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of them x and one of them
y and the height z.

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There.

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So if we give the box these
dimensions x, y, and z, we

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need to ask, well first of
all, what is its volume?

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And second of all, how much
cardboard is used?

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And as long as we can keep the
volume equal to 3, what's the

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least amount of cardboard
we can use?

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So the volume of this box is
just xyz and that has to be 3.

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So we know xyz is equal to the
volume, which is equal to 3.

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So we have a constraint
on x, y, and z here.

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And the thing that we
want to optimize--

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that we want to find
a minimum for--

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is the total amount
of cardboard used.

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So let's talk about how much
cardboard is used.

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So the bottom of the box
is triple-thick.

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So its area is x times y.

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So the total amount of cardboard
used is 3xy.

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Now, the front and back of the
box are single thickness.

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And they have area
xy, but there--

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sorry, not xy, xz, right?

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This height is z, so it's
x times z-- and

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there are two of them.

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So we have xz total cardboard
in front and xz total

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cardboard in back.

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Oh, I guess this has another
segment there in back because

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it's an open box.

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So xz and xz.

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So that's 2xz coming from
the front and back.

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And from the two sides, well
each side has area yz.

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There are two of them and
they're double-thick.

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So that contributes 4yz
to the total amount

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of cardboard used.

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So this is the total amount
of cardboard used.

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And we know xyz is equal to 3.

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So we want to minimize this, but
we want to minimize this

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taking this into account, taking
this restriction on the

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volume into account.

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So what we can do is you can
realize that we can just

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eliminate one of these
variables.

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Right?

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We can say z is equal
to 3 over xy.

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That always has to be true.

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And if we make that substitution
in here, then

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we'll be able to minimize this
expression without regard to

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any constraint anymore.

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So from xyz equals 3, we have
z equals 3 over xy.

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So we want to minimize what we
get when we plug z equals 3

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over xy into here.

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So that's 3xy, plus--

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well, 2x times 3 over xy is--

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6 over y, plus-- and 4y
times 3 over xy is--

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12 over x.

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And one other thing I guess we
haven't mentioned explicitly

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is that this is a
physical box.

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It has actual dimensions, so its
dimensions have to be, you

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know-- they're lengths, they
have to be positive numbers.

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So we want for x to be positive
and we want for y to

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be positive numbers.

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OK.

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So now we've got
this function.

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And there aren't--

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x and y can be anything, any
value of x and y we choose.

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This describes the amount of
cardboard used in a box with

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those dimensions on its base
that has volume 3.

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All right?

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So we want to minimize that.

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So now we've finally got to
the point of our calculus.

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Let's call this function--

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let's give it a name like,
I don't know--

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f of x, y.

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So we want to find the
global minimum of f.

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We want to find the absolute
least amount of cardboard that

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we can use.

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So there are several
possibilities for where a

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global minimum can occur.

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It can occur on some critical
point of the function, or it

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can occur on the boundary of the
region where the function

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is defined.

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And in this case, that includes
the possibility that

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it can occur as x or y or
both go off to infinity.

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All right?

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So we have to look at
those possibilities.

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In particular, we have to look
at the critical points.

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That's one of the
possibilities.

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So the critical points of f.

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So we need to find out what
those points are.

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So in order to find the
critical points,

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well what do we do?

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We do the usual thing.

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We look at its partial
derivatives.

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So we need the first and
second-- or sorry--

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the first partial derivatives
with respect to x and y, we

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need both of them to
be equal to 0.

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So we need fx equals
0, and fy equals 0.

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OK, so let's write down what
that means over here.

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So what is fx?

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Well, we just take
the partials.

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So it's 3y--

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plus we take the partial of 6y
with respect to x and we get

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0, and we take the partial of 12
over x with respect to x--

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and we get minus 12
over x squared.

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So that's fx, and we need
that to be equal to 0.

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And fy is equal to 3x minus 6
over y squared, and we also

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need that to be equal to 0.

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So if we solve this first
equation, for example, we can

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solve it for y in terms of x.

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And that tells me that
y is equal to--

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I need-- let's see--

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12 over x squared divided
by 3, so that's--

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4 over x squared.

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Now I can plug this 4 over x
squared in down here, and so I

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get 3x minus 6.

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Well, 6 divided by 4 divided by
x squared quantity squared

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is 6 times x squared
over 4 squared.

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That has to be equal to 0.

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And I can-- let's see-
I can rewrite this.

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Maybe I'll divide through
by 3 as that 3

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isn't going to matter.

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So it's x minus--

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so then I'll be left with 2
times x to the fourth over 16,

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so that's--

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x to the fourth over
8 equals 0.

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And you can see in this equation
that either x is

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equal to 0, or we can divide
by x and then we get x

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cubed equals 8.

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So x equals 0 or x
cubed equals 8.

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The only solution
is x equals 2.

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Now, it's easy to see that we
can't have a box with x equal

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to 0, right?

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Our function is not actually
defined at x equals 0 over

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here, you'll see, right.

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Our function had a 12 over x
in it, so we can't have x

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equal to 0.

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So that's not going to be
a critical point of this

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function. x equals 0 isn't going
to lead to a critical

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point, so, OK, so our only
critical points are going to

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happen when x is equal to 2.

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And when x is equal to 2, we can
go back up here and we see

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that y was equal to 4
over x squared at

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our critical points.

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So the only critical
point is 2, 1.

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OK.

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So that's our critical point.

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So now, we want to think about
is this point going to be a

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maximum or a minimum or what?

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And where is the global
minimum of this

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function going to occur?

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So remember that the global
minimum of this function can

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occur either on the critical
points or on the boundary or

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at a point where x or y
is going to infinity--

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in the limit, I guess I mean.

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So lets go back here and take
a look at our function here.

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So we can see from the
expression for this function

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that if x and y are going to
infinity, this is no good.

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This part is going to
go to infinity.

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3xy is going to go to infinity,
and these are going

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to be positive, so f is going
to go to infinity.

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OK, so as the dimensions of our
box get very, very large,

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the total amount of cardboard
used becomes infinite.

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Also, if x or y gets closer
and closer to 0--

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that's at the boundary--

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well, if x gets closer and
closer to 0, than 12 over x

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gets closer and closer
to infinity.

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00:11:31,240 --> 00:11:35,020
And if y gets closer and closer
to 0, then 6 over y

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gets closer and closer
to infinity.

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And again, all these terms are
going to be positive because x

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and y have to be positive.

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So in those limiting cases, the
function f of x, y gets

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infinitely large.

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So along the boundary and as x
and y go to infinity, this

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function blows up.

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It gets very, very big.

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So that means the only place it
could have a global-- so it

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has a global minimum-- the only
place that global minimum

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can be is at a critical point.

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00:12:01,080 --> 00:12:04,230
And we saw, if we walk back over
here again, that the only

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critical point is
the point 2, 1.

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All right?

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So that means this point
has to be a global

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minimum for the function.

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Now, you might, out of
curiosity, you might ask how

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much cardboard are we actually
talking about in that case?

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Since that's really the one
part, I guess wan't really

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part of the question, but its
an interesting thing to ask.

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00:12:25,310 --> 00:12:35,110
So in this case, we use f of 2,
1 equals-- well let's see--

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so that's 6 plus 6
plus 6 equals 18

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units squared of cardboard.

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Now, I guess one thing you might
notice is we didn't use

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the second derivative
test here.

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And the question is why--

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I mean, we concluded it was a
global maximum without ever

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using the second derivative
test. Sorry, global minimum.

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Pardon me.

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00:13:03,730 --> 00:13:05,530
And the second derivative test
is the thing that tells us

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whether things are minimum or
maximum or saddle points.

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So why didn't we use it?

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Well, the answer is the second
derivative test tells you

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whether something is a local
minimum or maximum or whether

250
00:13:16,070 --> 00:13:18,220
it's a saddle point.

251
00:13:18,220 --> 00:13:20,760
So if we had applied the second
derivative test at this

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00:13:20,760 --> 00:13:23,050
critical point, what we would
have learned is that this

253
00:13:23,050 --> 00:13:25,480
critical point is
a local minimum.

254
00:13:25,480 --> 00:13:28,700
But being a local minimum isn't
enough to guarantee that

255
00:13:28,700 --> 00:13:29,610
it's a global minimum.

256
00:13:29,610 --> 00:13:31,600
Because we could have
on the boundary--

257
00:13:31,600 --> 00:13:33,490
or as x or y goes
to infinity--

258
00:13:33,490 --> 00:13:35,620
we could have that the function
value get smaller and

259
00:13:35,620 --> 00:13:36,830
smaller without bound.

260
00:13:36,830 --> 00:13:38,340
Now in this case, we
showed that the

261
00:13:38,340 --> 00:13:39,720
function doesn't do that.

262
00:13:39,720 --> 00:13:41,660
It gets larger and larger
without bound.

263
00:13:41,660 --> 00:13:45,450
And so that meant that that
minimum point really is the

264
00:13:45,450 --> 00:13:47,430
global minimum.

265
00:13:47,430 --> 00:13:51,360
But the second derivative test
isn't enough to conclude that

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00:13:51,360 --> 00:13:53,030
something is a global
minimum on its own.

267
00:13:53,030 --> 00:13:56,600
You really do need that extra
analysis that we did.

268
00:13:56,600 --> 00:13:58,220
I'll end there.

269
00:13:58,220 --> 00:13:58,372