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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

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is how to solve some
max-min problems.

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How to find the maximum
or minimum of a given

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function given some constraints
and figure out where it reaches

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its largest 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

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are going to be made just of
single-thickness cardboard.

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But the two sides,
the left and right,

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are going to be
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

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just going to be an open box.

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

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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,

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a lot of 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 of cardboard possible?

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All right?

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So we want the
dimensions of the box

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that use the least
total amount cardboard.

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

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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

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a little 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--
we want to figure out

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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

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them is we can call
one of them x and one

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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,

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we 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

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equal to 3, what's the least
amount of cardboard we can use?

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So the volume of this box is
just x*y*z and that has to be

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3.

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So we know x*y*z 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 3x*y.

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

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And they have area x*y, but
there-- sorry, not x*y, x*z,

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right?

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

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and there are two of them.

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So we have x*z total cardboard
in front and x*z total

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cardboard in back.

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Oh, I guess this has another
segment there in back

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because it's an open box.

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So x*z and x*z.

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So that's 2x*z coming
from the front and back.

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And from the two sides,
well each side has area y*z.

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

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So that contributes 4y*z to the
total amount of cardboard used.

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So this is the total
amount of cardboard used.

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And we know x*y*z is equal to 3.

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So we want to
minimize this, but we

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want to minimize
this taking this

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into account, taking this
restriction on the volume

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into account.

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So what we can do
is you can realize

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that we can just eliminate
one of these variables.

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Right?

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We can say z is
equal to 3 over x*y.

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That always has to be true.

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And if we make that
substitution in here,

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

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to any constraint anymore.

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So from x*y*z equals 3, we
have z equals 3 over x*y.

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

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over x*y into here.

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So that's 3x*y, plus-- well, 2x
times 3 over x*y is 6-- over y,

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plus-- and 4y times 3
over x*y is 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

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to be, you know--
they're lengths,

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they have to be
positive numbers.

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

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to 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-- x
and y can be anything,

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any value of x and y we choose.

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This describes the
amount of cardboard

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used in a box with
those dimensions

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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--
let's give it a name

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like, I don't know-- 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

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that we can use.

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So there are several
possibilities

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for where a global
minimum can occur.

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It can occur on some critical
point of the function,

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or it can occur on the
boundary of the region

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where the function is defined.

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And in this case, that
includes the possibility

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that 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, 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,

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we need both of them
to be equal to 0.

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So we need f_x equals
0, and f_y equals 0.

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

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So what is f_x?

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Well, we just take the partials.

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So it's 3y-- plus we take the
partial of 6y with respect

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to x and we get 0, and we
take the partial of 12 over x

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with respect to x-- and we
get minus 12 over x squared.

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

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And f_y is equal to 3x
minus 6 over y squared,

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and we also need that
to be equal to 0.

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So if we solve this first
equation, for example,

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we can 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-- 12
over x squared divided by 3,

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so that's 4 over x squared.

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

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and so I get 3x minus 6.

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Well, 6 divided by 4 divided
by x squared quantity

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squared 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

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that 3 isn't going to matter.

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So it's x minus-- so then
I'll be left with 2 times x

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to the fourth over 16, so
that's-- x to the fourth over 8

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equals 0.

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And you can see in this equation
that either x is equal to 0,

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or we can divide by x and
then we get x 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

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a box with x equal 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,

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so we can't have x equal to 0.

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So that's not going
to be a critical point

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of this function.
x equals 0 isn't

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going to lead to a
critical point, so, OK,

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so our only critical
points are going

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to happen when x is equal to 2.

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

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and we see that y
was equal to 4 over x

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squared at 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

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this point going to be a
maximum or a minimum or what?

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And where is the global
minimum of this function

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going to occur?

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So remember that the global
minimum of this function

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can occur either on the critical
points or on the boundary

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or 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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3x*y is going to go to infinity,
and these are going to be

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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

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to 0-- that's at
the boundary-- well,

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if x gets closer
and closer to 0,

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than 12 over x gets closer
and closer to infinity.

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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

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to be positive because x
and y have to be positive.

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

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gets infinitely large.

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

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this 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 can

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be is at a critical point.

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00:12:01,080 --> 00:12:03,190
And we saw, if we
walk back over here

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again, that the only 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 minimum

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for the function.

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Now, you might,
out of curiosity,

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you might ask how much cardboard
are we actually talking

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about in that case?

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Since that's really
the one part--

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I guess it wasn't really
part of the question,

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but it's an interesting
thing to ask.

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So in this case, we use
f of 2, 1 equals-- well,

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let's see-- so that's 6 plus 6
plus 6 equals 18 units squared

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of cardboard.

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Now, I guess one
thing you might notice

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is we didn't use the second
derivative test here.

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And the question
is why-- I mean,

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we concluded it was a global
maximum without ever using

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the second derivative test.

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Sorry, global minimum.

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Pardon me.

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And the second derivative
test is the thing

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that tells us whether
things are minimum

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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

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tells you whether something
is a local minimum or maximum

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or whether it's a saddle point.

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So if we had applied the
second derivative test

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at this critical point,
what we would have learned

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is that this critical
point is a local minimum.

245
00:13:25,480 --> 00:13:28,486
But being a local minimum
isn't enough to guarantee

246
00:13:28,486 --> 00:13:29,610
that it's a global minimum.

247
00:13:29,610 --> 00:13:31,600
Because we could have
on the boundary--

248
00:13:31,600 --> 00:13:33,544
or as x or y goes
to infinity-- we

249
00:13:33,544 --> 00:13:35,960
could have that the function
value get smaller and smaller

250
00:13:35,960 --> 00:13:36,830
without bound.

251
00:13:36,830 --> 00:13:39,720
Now in this case, we showed that
the function doesn't do that.

252
00:13:39,720 --> 00:13:41,660
It gets larger and
larger without bound.

253
00:13:41,660 --> 00:13:45,170
And so that meant that
that minimum point really

254
00:13:45,170 --> 00:13:47,430
is the global minimum.

255
00:13:47,430 --> 00:13:49,300
But the second
derivative test isn't

256
00:13:49,300 --> 00:13:51,690
enough to conclude
that something

257
00:13:51,690 --> 00:13:53,030
is a global minimum on its own.

258
00:13:53,030 --> 00:13:56,600
You really do need that
extra analysis that we did.

259
00:13:56,600 --> 00:13:57,872
I'll end there.