WEBVTT
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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.
00:08:01.550 --> 00:08:09.490
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.
00:08:17.820 --> 00:08:22.490
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
00:09:13.250 --> 00:09:14.170
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
00:10:01.200 --> 00:10:04.240
and we see that y
was equal to 4 over x
00:10:04.240 --> 00:10:05.780
squared at our critical points.
00:10:05.780 --> 00:10:21.370
So the only critical
point is (2, 1).
00:10:21.370 --> 00:10:22.210
OK.
00:10:22.210 --> 00:10:23.700
So that's our critical point.
00:10:23.700 --> 00:10:26.310
So now, we want
to think about is
00:10:26.310 --> 00:10:31.730
this point going to be a
maximum or a minimum or what?
00:10:31.730 --> 00:10:36.010
And where is the global
minimum of this function
00:10:36.010 --> 00:10:37.300
going to occur?
00:10:37.300 --> 00:10:39.900
So remember that the global
minimum of this function
00:10:39.900 --> 00:10:44.170
can occur either on the critical
points or on the boundary
00:10:44.170 --> 00:10:49.790
or at a point where x or
y is going to infinity--
00:10:49.790 --> 00:10:51.470
in the limit, I guess I mean.
00:10:51.470 --> 00:10:55.300
So lets go back here and take
a look at our function here.
00:10:55.300 --> 00:10:59.590
So we can see from the
expression for this function
00:10:59.590 --> 00:11:03.270
that if x and y are going to
infinity, this is no good.
00:11:03.270 --> 00:11:05.030
This part is going
to go to infinity.
00:11:05.030 --> 00:11:07.446
3x*y is going to go to infinity,
and these are going to be
00:11:07.446 --> 00:11:10.160
positive, so f is going
to go to infinity.
00:11:10.160 --> 00:11:14.980
OK, so as the dimensions of
our box get very, very large,
00:11:14.980 --> 00:11:19.010
the total amount of cardboard
used becomes infinite.
00:11:19.010 --> 00:11:22.140
Also, if x or y gets
closer and closer
00:11:22.140 --> 00:11:25.230
to 0-- that's at
the boundary-- well,
00:11:25.230 --> 00:11:27.020
if x gets closer
and closer to 0,
00:11:27.020 --> 00:11:31.240
than 12 over x gets closer
and closer to infinity.
00:11:31.240 --> 00:11:35.020
And if y gets closer and
closer to 0, then 6 over y
00:11:35.020 --> 00:11:36.530
gets closer and
closer to infinity.
00:11:36.530 --> 00:11:38.030
And again, all these
terms are going
00:11:38.030 --> 00:11:40.210
to be positive because x
and y have to be positive.
00:11:40.210 --> 00:11:45.180
So in those limiting cases,
the function f of x, y
00:11:45.180 --> 00:11:46.890
gets infinitely large.
00:11:46.890 --> 00:11:49.635
So along the boundary and
as x and y go to infinity,
00:11:49.635 --> 00:11:50.940
this function blows up.
00:11:50.940 --> 00:11:52.160
It gets very, very big.
00:11:52.160 --> 00:11:55.010
So that means the only place
it could have a global-- so it
00:11:55.010 --> 00:11:58.850
has a global minimum-- the only
place that global minimum can
00:11:58.850 --> 00:12:01.080
be is at a critical point.
00:12:01.080 --> 00:12:03.190
And we saw, if we
walk back over here
00:12:03.190 --> 00:12:07.240
again, that the only critical
point is the point (2, 1).
00:12:07.240 --> 00:12:07.740
All right?
00:12:07.740 --> 00:12:09.870
So that means this point
has to be a global minimum
00:12:09.870 --> 00:12:10.910
for the function.
00:12:10.910 --> 00:12:12.617
Now, you might,
out of curiosity,
00:12:12.617 --> 00:12:14.950
you might ask how much cardboard
are we actually talking
00:12:14.950 --> 00:12:17.170
about in that case?
00:12:17.170 --> 00:12:19.901
Since that's really
the one part--
00:12:19.901 --> 00:12:22.710
I guess it wasn't really
part of the question,
00:12:22.710 --> 00:12:25.310
but it's an interesting
thing to ask.
00:12:25.310 --> 00:12:35.560
So in this case, we use
f of 2, 1 equals-- well,
00:12:35.560 --> 00:12:44.180
let's see-- so that's 6 plus 6
plus 6 equals 18 units squared
00:12:44.180 --> 00:12:46.560
of cardboard.
00:12:46.560 --> 00:12:49.910
Now, I guess one
thing you might notice
00:12:49.910 --> 00:12:52.400
is we didn't use the second
derivative test here.
00:12:52.400 --> 00:12:54.760
And the question
is why-- I mean,
00:12:54.760 --> 00:12:58.490
we concluded it was a global
maximum without ever using
00:12:58.490 --> 00:13:01.200
the second derivative test.
00:13:01.200 --> 00:13:02.250
Sorry, global minimum.
00:13:02.250 --> 00:13:03.269
Pardon me.
00:13:03.269 --> 00:13:05.060
And the second derivative
test is the thing
00:13:05.060 --> 00:13:06.870
that tells us whether
things are minimum
00:13:06.870 --> 00:13:08.036
or maximum or saddle points.
00:13:08.036 --> 00:13:09.260
So why didn't we use it?
00:13:09.260 --> 00:13:11.200
Well, the answer is the
second derivative test
00:13:11.200 --> 00:13:15.630
tells you whether something
is a local minimum or maximum
00:13:15.630 --> 00:13:18.220
or whether it's a saddle point.
00:13:18.220 --> 00:13:20.440
So if we had applied the
second derivative test
00:13:20.440 --> 00:13:22.610
at this critical point,
what we would have learned
00:13:22.610 --> 00:13:25.480
is that this critical
point is a local minimum.
00:13:25.480 --> 00:13:28.486
But being a local minimum
isn't enough to guarantee
00:13:28.486 --> 00:13:29.610
that it's a global minimum.
00:13:29.610 --> 00:13:31.600
Because we could have
on the boundary--
00:13:31.600 --> 00:13:33.544
or as x or y goes
to infinity-- we
00:13:33.544 --> 00:13:35.960
could have that the function
value get smaller and smaller
00:13:35.960 --> 00:13:36.830
without bound.
00:13:36.830 --> 00:13:39.720
Now in this case, we showed that
the function doesn't do that.
00:13:39.720 --> 00:13:41.660
It gets larger and
larger without bound.
00:13:41.660 --> 00:13:45.170
And so that meant that
that minimum point really
00:13:45.170 --> 00:13:47.430
is the global minimum.
00:13:47.430 --> 00:13:49.300
But the second
derivative test isn't
00:13:49.300 --> 00:13:51.690
enough to conclude
that something
00:13:51.690 --> 00:13:53.030
is a global minimum on its own.
00:13:53.030 --> 00:13:56.600
You really do need that
extra analysis that we did.
00:13:56.600 --> 00:13:57.872
I'll end there.