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[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. .