WEBVTT

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PROFESSOR: Welcome
back to recitation.

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This'll be the last video where
we do an 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--
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,

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I'm going to remind you of
the two formulas for volume

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and 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*r*h.

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So with that information,
I'll give you

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some time to work on 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

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trying to solve an
optimization problem.

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And so we know now the
constraint equation,

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it's a little different
than the other situations

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because the
constraint equation is

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just we're told that
there's a fixed volume

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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

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about exact values
for radius and height,

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but what ratio between them
will minimize surface area.

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

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things will look a
little different.

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But ultimately we still have
our optimizing equation,

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which is surface area.

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And we still have our constraint
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

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and use our constraint equation
to get rid of a variable

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so that we can write
the right-hand side here

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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

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of just a function of r.

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

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didn't think of doing,
but ultimately you

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should end up with the same
answer once you've simplified.

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Notice that this
term has a pi*r*h.

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This term also has a pi*r*h.

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

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equal to pi*r*h.

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So what I'm going to do is take
that pi*r*h here and replace it

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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

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that I do when 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

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from the problem
I see, oh, there's

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something here
that looks exactly

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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,

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but it's maybe a little faster.

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So now the new surface area
equation becomes 2*pi r squared

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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,

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but this term goes to infinity.

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So as r gets very large surface
area is getting very large.

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As r goes to 0 this
term goes to 0,

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but this 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

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a derivative with
respect to radius

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it's going to have
to be a minimum.

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So we don't have to check
any more-- now we've

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checked sort of the what's
happening towards the boundary

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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

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and find 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

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I have r 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 keep the 2V,

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and the denominator 1 over r,
its derivative is negative 1

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over r squared.

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

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2V over r squared.

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

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

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which is the same as-- going
to put it on this side--

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r cubed is equal to 2*pi
over 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.

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Let me double 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.

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I 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 video,

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scary things can happen.

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2V-- 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 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 sides

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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

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and see if we can figure 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,

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I'll end up with 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

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

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Which is, by the
way, the same thing

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as multiplying by
pi 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 height

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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

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just asks, what is the ratio
between radius and height

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that 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--
I divided by r squared--

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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

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the same thing that I do on the
left-- I divided by V over pi,

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which is multiplying by pi
over V. The pi's divide out,

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the V's divide out,
and I'm left with 1/2.

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And so this was an optimization
problem where the constraint

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

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it did have a fixed constant.

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

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and an exact value
for the height,

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but I could ask you
for how 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.