[music].
Let's suppose that you work for a soup
company, and they've given you a task, a
task to design a soup can that can hold
300 times pi cubic centimeters of soup.
This is really an optimization problem.
Your goal is to design a soup can using as
little metal as possible, which
nevertheless holds 300 pi cubic
centimeters of soup.
Okay, less dramatically, you're really
being asked to produce a cylinder with a
given volume.
I want the volume of the thing to be 300
pi cubic centimeters, and I want you to
minimize the surface area, right.
This thing's got a side and a top and a
bottom.
And I want you to minimize the surface
area for a given volume.
Probably helps to review a little bit of
the geometry that goes into this problem.
Here's a picture of a cylinder right, and
it's got some height that I'll be calling
h and a radius that I'll call r.
And the volume of this cylinder is pi r
squared h.
All right, pi r squared is the area, and
I'm multiplying by h to compute the volume
of this disk dragged through space.
What's the surface area of this cylinder?
Well 2 pi r squared counts the surface
area of the top disk and the bottom disk.
Each of those has pi r squared for their
area.
And the curved part of this cylinder has
area 2 pi r h.
Okay, let me draw a picture of my soup
can.
So, yeah, here's a little picture of our
soup can, and you see I've labeled the
relevant sizes.
The height of the can and the radius of
the can is what I'm trying to figure out.
Now what is it that I'm trying to
optimize?
So I'm trying to minimize the surface
area.
Here's that formula for the surface area
of this can, and I want that surface area
to be as small as possible.
Of course, if I wanted to make that
surface area as small as possible, I just
make a really tiny soup can, right?
But there's a constraint.
Well, the constraint is that soup can has
to hold 300 pi cubic centimeters of soup.
So I'm trying to make this quantity, the
surface area, as small as possible,
subject to the constraint that the volume
of the soup can is in fact 300 pi.
And I should also point out that the
radius and height had better be positive.
Otherwise, kind of nonsense.
This thing that I'm trying to optimize
really involves two variables.
It involves the height and also a radius.
Yeah this is some bad news, right?
The quantity that I'm trying to minimize
involves 2 variables, involves r and h.
And calculus, as we've been setting it up,
only works for understanding how a single
variable changing affects something else.
So I need to rewrite this function of two
variables as a function of a single
variable.
So, here we go, let's take this constraint
and let's solve for h, all right?
And if I start doing that, well, h is 300
pi over pi r squared and I can cancel
these, pis and I'm just left with h is 300
over r squared.
So, for a given radius, this is how tall I
need to make the soup can to guarantee
that the soup can holds 300 pi cubic
centimeters of soup.
Now, once I've got an equation for h in
terms of r, I can use this to rewrite the
thing I'm trying minimize just in terms of
r.
So, here we go.
The thing I'm trying to minimize now is 2
pi r squared plus 2 pi r times h, but h,
in order to satisfy the constraint, is 300
over r squared.
And now I can simplify this a little bit
further, right?
I've got an r and an r squared here.
So I can get rid of this square and get
rid of this r.
And then, the thing I'm trying to minimize
is 2 pi r squared plus 2 pi times 300,
which is 600 pi divided by r.
So this is the quantity that I'm trying to
minimize.
Now that I've got this thing written as a
function of a single variable, I'm really
tempted to do the fifth step here.
Apply calculus, right?
If I got a function of a single variable,
I could differentiate.
Find the critical points.
Figure out where the maxima and minima
occur.
But before we dive in to the calculus side
of this problem, let's just think about
some creative solutions to the soup can
issue.
For example, here's a little model of a
situation.
Here's the sun, here's Earth and here is
my new plan for soup cans, right?
Really big radius, really tiny height,
right?
Even if r is you know, 100 million miles,
that can make h small enough so that this
soup can contains exactly 300 pi cubic
centimeters of soup.
[laugh].
Well, what's the surface area in this
case?
Is this a really good choice for a soup
can, in, in terms of surface area?
No, right?
If I make r really big, admittedly, this
quantity, 600 pi over r, which is
measuring the curved area, that'll end up
being very small.
Right, but 2 pi r squared when r is really
big, this quantity, the amount of metal in
the top and the bottom of the can is
enormous right?
So this is not a great choice for soup
cans.
So having the soup can shaped like a giant
pancake is a terrible idea.
But what if I went the other way?
Here is a piece of wire, what if I made
the soup can really long and thin like
this piece of wire?
Well, here is that long, thin wire-shaped
soup can, right.
Even if I make the radius of my cylinder
very small, if I make it long enough, then
the soup can does hold 300 pi cubic
centimeters of soup, right?
And the question is, is this a really
smart choice to minimize surface area, and
no, it turns that if r is really small,
that's really not all that helpful.
Yeah, if r is really small, then 2 pi r
squared is really small.
I don't need very much metal on the top
and the bottom of the soup can.
But then 600 pi over r, which is how much
metal is in the curved part of the soup
can, this quantity is going to be
enormous.
So really, we see that these two things
are competing against each other.
A small value of r isn't really very
helpful, and a really big value of r isn't
very helpful.
So our creative solutions didn't quite pan
out.
Let's do some calculus now.
Okay, so I'm going to think of this
surface area as a function of a single
variable, r, right?
Here it is, f of r is 2 pi r squared plus
600 pi over r, and I'm going to
differentiate, and here's the derivative.
All right, I differentiate 2 pi r squared
to get 4 pi r, and I differentiate this
number divided by r.
Well, 1 over r is negative 1 over r
squared.
That's its derivative.
And the I multiply by 600 pi.
So this, is the derivative of this.
Now, I need to find the critical points.
Now, remember what critical points are,
right?
Critical points are where the derivative
vanishes, or where the derivative doesn't
exist, right?
The function's not differentible.
This function, that was differentible on
its domain.
All right?
You might worry when r is equal to 0.
But that's not even in the domain of the
original function.
So I don't have to worry about that.
All right, now we've gotta figure out,
when is the derivative equal to zero, so
I'm trying to solve this.
I'll add 600 pi over r squared to both
sides.
So 4 pi r must be 600 pi over r squared in
that case.
I'll multiply both sides by r squared, and
I'll divide both sides by 4 pi.
And I'll get r cubed is 600 pi over 4 pi.
The pis cancel, and 600 over 4 is 150, so
r cubed must be 150.
In other words, r is the cube root of 150.
This is the only critical point.
Now that I found the critical points, what
about the end points?
Well, the end points actually aren't
really valid in this situation.
One of the end points might be when r is
equal to zero.
But in that case, the soup can has no
volume.
So I can't satisfy the constraint.
The same thing happens when h is equal to
zero.
In that case, the soup can, again, has no
volume.
So I can't satisfy the constraint.
So, I don't have to worry about the end
points, because they're not in the domain
that I'm considering.
But I do have to worry about the limiting
behavior when r is really big or when r is
really small.
So, we actually already handled the
limiting behavior.
Right?
We've already considered the situation
when r is really small.
That's the situation where the soup can's
like a long, thin wire.
And we checked, in that case the surface
area is enormous.
Right?
If r is small enough, the surface area can
be as large as you like, and still hold
300 pi cubic centimeters of soup.
We also consider the situation where r is
really large, right?
When r was really big ,we had this sort of
flat pancake shape for the soup can.
And in that case, we again saw that if r
is big enough, the surface area can be
made as large as you like, while still
having that soup can contain 300 pi cubic
centimeters of soup.
So, the only thing that really remains is
this critical point.
And if you plug this critical point into
the original function, you get this as the
resulting surface area, if you build your
soup can with a radius of cube root 150
centimeters, and your height, whatever it
has to be in order to guarantee that the
soup can contains 300 pi cubic centimeters
of soup.
So how does this turn out?
What's the best shape for a soup can?
So the best case for your soup can is to
use a radius of the cube root of 150.
And you can see that from this graph.
Here, I'm graphing the surface area for a
given radius of soup can containing 300 pi
cubic centimeters of soup.
And you can see that the graph goes up
towards infinity here.
And up towards infinity there.
And that's exactly the situation when the
radius is very small or very big.
The derivative is negative and then
positive.
Right?
The function's decreasing and then
increasing.
So this value really is a local minimum.
And in fact, this is the global minimum of
this function.
Right?
And that's why this is the best choice for
your soup can.
Is this really a legitimate application of
calculus?
Well, here's a task to think about.
Right?
Go out, find some soup cans and check
them.
Are they really building soup cans in
order to minimize the amount of metal that
it takes to manufacture the can?
Or perhaps there's other issues involved
in manufacturing soup cans.
.