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PROFESSOR: Welcome back
to recitation.

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This'll be the last video
where we do an

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optimization problem.

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And this one's a little bit
different than the other two.

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So I'm going to give you
the problem now.

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The problem is the following--

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a cylinder has a fixed volume.

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What ratio between radius and
height minimizes surface area?

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Before I give you time to think
about that, I'm going to

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remind of the two formulas
for volume and

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surface area of a cylinder.

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So volume of a cylinder
is pi r squared h.

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And surface area is 2 pi
r squared plus 2 pi rh.

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So with that information, I'll
give you some time to work on

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this problem and then
I'll be back.

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Welcome back.

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OK, what we're doing, again,
is we're trying to solve an

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optimization problem.

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And so we know now the
constraint equation, it's a

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little different than the other
situations because the

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constraint equation is just
we're told that there's a

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fixed volume but we're
not told what it is.

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That actually will not change
how we work on this problem,

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but it does change the kind of
answer I can ask of you.

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And notice the answer I ask is
not about exact values for

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radius and height, but what
ratio between them will

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minimize surface area.

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So this is, you'll see at
the end, things will

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look a little different.

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But ultimately we still have our
optimizing equation, which

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is surface area.

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And we still have
our constraint

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equation, which is volume.

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So again, we're going to
do what we always do.

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We're going to take our
optimization equation and use

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our constraint equation to get
rid of a variable so that we

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can write the right hand side
here in terms of one variable.

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Let me point out V is
not a variable here.

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V is a constant because
the volume is fixed.

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So when you see V, it's
not a variable.

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It's a constant.

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So what do I do?

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Let me write, first, let me
write surface area in terms of

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just a function of r.

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And I'm going to do something
a little tricky which maybe

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you didn't think of doing, but
ultimately you should end up

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with the same answer once
you've simplified.

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Notice that this term
has a pi rh.

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This term also has a pi rh.

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In fact, I can rewrite the
volume equation as V over r is

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equal to pi rh.

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So what I'm going to do is
take that pi rh here and

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replace it by a V over r.

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Now again, you might
not have done this.

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You should ultimately get the
same answer that I do when

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we're finished.

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And even sooner, probably
some simplification

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would be the same.

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The only thing I've done here is
I've simplified right away.

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So by looking at the problem and
kind of pulling back from

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the problem I see, oh, there's
something here that looks

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exactly like something
over here.

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So I just want to
point that out.

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That it's not wrong to just
substitute for h, but it's

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maybe a little faster.

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So now the new surface area
equation becomes 2 pi r

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squared plus 2V over r.

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So now we have everything in
terms of r, because again, V

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is a constant.

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Now let me point out
what happens.

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When r goes to infinity this
term goes to 0, but this term

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goes to infinity.

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So as r gets very large
service area is

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getting very large.

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As r goes to 0 this term
goes to 0, but this

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term goes to infinity.

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V is fixed, and when r goes
to 0 this term blows up.

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So when r gets as small as we
allow or as large as we allow,

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either way, surface area is
going to be getting big.

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So where this, where surface
area has a derivative with

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respect to radius it's going
to have to be a minimum.

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So we don't have to
check anymore.

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Now we've checked sort of the
what's happening towards the

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boundary at the extreme
values of r.

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So now we can, now we can
actually solve the problem.

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Well, what do we do?

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We're using our optimization
equation.

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We want to take a derivative and
set it equal to 0 and find

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what value for r gives that.

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But again, let me point out one
more time that in the end

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I'm asking for a ratio between
radius and height.

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So I'm not going to get all
the way to where I have r

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equals something.

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You'll see I'm going to
do another trick.

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But let me first take the
derivative I need.

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Surface area prime, this is
derivative with respect to r.

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I have 2 pi r squared.

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That derivative with respect
to r is 4 pi r.

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And this derivative with respect
to r, I'm going to

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keep the 2V, and the denominator
1 over r, its

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derivative is negative
1 over r squared.

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So I'm going to have 4 pi r for
the first term and then

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minus 2V over r squared.

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If I set this equal to 0, the
derivative equal to 0 and

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solve, I get to 2V over r
squared is equal to 4 pi r,

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which is the same as--

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going to put it on this side--
r cubed is equal to 2 pi over

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V. Let's just check.

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If I multiply both sides by
r squared, divide by 4 pi.

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Oh, I did it backwards.

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I think it should actually be
pi over 2V. Let me double

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check that all my signs
are correct.

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That's r cubed, divide by 4 pi,
I should get pi over 2V. I

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should not.

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I've been told from the audience
I made a mistake.

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Sometimes you'll see when you're
at the board and on

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video, scary things
can happen.

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2V--

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oh.

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Multiply through by
r, I get r cubed.

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Oh, I have the 2V in
the numerator.

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I apologize.

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The 2V is in the numerator,
the 4 pi is in the

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denominator.

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So I get V over 2 pi.

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Does that look better,
audience?

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The audience tells me
that looks better.

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OK.

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So here I am.

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I have r cubed is equal
to V over 2 pi.

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Now, if I wanted to I could
take the cube root of both

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sides and get r explicitly
in terms of V and 2 pi.

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V is a constant, pi is a
constant, I would be done.

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But I didn't ask for
what r actually is.

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I asked for a ratio.

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So let's make this
problem simpler.

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All I need in the end
is r divided by h.

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Let's go back to a formula we
have and see if we can figure

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out a way to get
r divided by h.

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Look at this formula for
V. It has an r squared

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and it has an h.

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So if I divide by r squared h on
this side, I'll end up with

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an r over h.

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Hopefully you buy that.

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I'm going to even
write it out.

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I'm going to take r cubed and
divide it by r squared h.

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That in the end, is r over h.

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Let's go back here look
at what that equals.

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r squared h is equal
to V over pi.

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Right?

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r squared h is equal
to V over pi.

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So I can divide this side by r
squared h and divide this side

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by V over pi and it's
the same thing.

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Which is, by the way, the same
thing as multiplying by pi

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over V. Right?

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So I can multiply this side by
pi over V, and what do I get?

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I get the V's divide out, the
pi's divided by, divide out,

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and I get 1/2.

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So the result is that the
ratio between radius and

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height should be 1 to 2.

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Let me one more time explain
what we were doing here.

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In the end, the answer, the
question just asks, what is

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the ratio between radius
and height that

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minimizes surface area?

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And I had an r cubed, I wanted
to get r divided by h.

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So I divided by two r's--

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I divided by r squared--
and then h again.

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But r squared h is V over pi, so
if I divide by r squared h

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I'm dividing by V over pi.

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So I can do on the right hand
side the same thing that I do

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on the left--

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I divided by V over pi, which
is multiplying by pi over V.

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The pi's divide out, the
V's divide out, and

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I'm left with 1/2.

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And so this was an optimization
problem where the

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constraint equation did not have
a number in it, but it

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did have, it did have
a fixed constant.

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So I couldn't ask you for an
exact value for the radius and

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an exact value for the height,
but I could ask you for how

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they relate.

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And that's ultimately
what I did.

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And I think that's
where we'll stop.

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