Here's another timeless classic, another
optimization problem that just comes up in
all the calculus textbooks.
Goes like this: you've got some big object
and you want to move that big object
around a corner.
The question is for some given corner, how
big of a object can you navigate around
that corner.
This is an example of the so called piano
moving problem.
A piano is way to complicated of an object
to really understand in this course right,
it's just got a way to much geometry going
on there.
So going to consider a simpler example.
So here's our diagram of a corner and I
mentioned that I want to move this red
stick around this corner without it
getting stuck.
Maybe it doesn't look like it but this
problem can be set up as an optimization
problem and we can solve it.
It's an optimization problem, because I'm
asking to know the length of the longest
stick, that I can navigate around this
corner.
We have to think about where that stick
can get stuck.
Well the stick is going to get stuck when
it's in a position like this.
When it's touching these two walls and
pressed up against this corner.
So, what I actually have to think about is
what's the shortest length of stick that
simultaneously touches this wall, this
wall and this corner.
I think it's a little bit funny that in
order to solve this problem, we end up
minimizing something.
We actually want to figure out the longest
length, the maximum length of a stick that
can be navigated around that corner.
But to answer that problem, we end up
solving a minimization problem.
Well anyhow, let's proceed in our usual
way.
I'm going to draw a picture and label
everything.
So, here I have drawn this picture and I
have labelled every thing.
The width of this hallway is a, the width
of this hallway is b, the length of this
stick is l, and makes this angle theta
with this bottom wall.
Now we gotta figure out what it is that we
are trying to optimize.
What's the quantity that we are trying to
minimize?
What's the goal?
So my goal is to minimize the length of
this stick, right?
I want to know the shortest length that
touches these two walls and this corner,
and that'll constrain the largest stick
that'll fit around this corner.
Now, my hallways have fixed lengths, and
I'd like to express the length of this
stick in terms of this angle theta.
So I can break up the stick into 2 pieces.
There's a piece that goes from this wall
to the corner and a piece from this corner
to the other wall.
So this piece here, well that's really the
hypotenuse of a right triangle, whose
opposite side to this angle theta has
length a.
And that means this hypotenuse has length
a times co-secant theta.
Similarly here I got a right triangle and
I have got this side length here is b so,
the hypotenuse has length b times secant
theta and the stick is just the sum of
these two lengths so that gives me a
formula for l in terms of theta.
L is a cosecant theta plus b secant theta.
What's the constraint in this problem?
Yeah I only need to think about theta
being between 0 and pi over 2 in order to
touch these two walls and the corner.
So now we've got a function, and it's
function of a single variable so we can
apply calculus.
So, I'm going to differentiate this
function, right, the derivative of L with
respect to theta.
Well what's the derivative of cosecant?
That's minus cosecant theta cotangent
theta.
And what's the derivative of secant?
Well that's secant theta Tangent theta.
So that's the derivative of l minus a
cosecant theta cotangent theta plus b
secant theta tangent theta.
With the derivative in hand I can now try
to find the critical points for this
function.
Now since this function L is
differentiable on the domain that we're
considering, I don't have to worry about
places where L fails to be differentiable
when I'm looking for critical points.
I only need to find points where the
derivative is equal to zero.
So for which values of theta does this
thing vanish?
Well, let's write this as an equation
equal to 0, and let's add a cosecant theta
contangent theta to both sides.
So then I've got b secant theta tangent
theta is equal to a cosecant theta
cotangent theta.
I, I will divide both sides by co-secant
theta, cotangent theta and I will divide
both sides by b so that I have got secant
theta, tangent theta divided by cosecant
theta, cotangent theta is equal to a over
b.
Now what do I notice here?
Well dividing by cotangent is as good as
multiplying by tangent.
And secant and co-secant is actually
tangent again.
So this is in fact tangent cubed.
So I've got tangent cubed theta.
Is equal to A over B.
If I take a cubed route now, I get that
tangent theta is equal to the cube root of
A over B.
In over words, theta is arc tangent the
cube root of A over B.
So now I know what theta is, but I don't
care about theta.
What I really care about is L.
What's the length?
So let's evaluate l of theta at the
critical point.
Now, where's the critical point?
It's where tangent of theta is equal to
the cube root of a over b.
I can build a right triangle with an angle
whose tangent is the cube root of a over
b.
I do that as follows.
I make this length 1.
And this length the cube root of a over b
and that makes the tangent of theta which
is this divided by this equal to this
right and this is 1 in the denominator.
That is the right triangle which means I
can compute the length of hypotenuse
right.
The length of hypotenuse is a square root
of the sum of these two lengths squared
and that's what I had written.
And here.
Okay, so I've got a right triangle.
I've got an angle whose tangent is theta
and I'm trying to calculate L of theta.
And if you remember back L was defined in
terms of cosecant and secant.
And I can compute cosecant and secant for
this angle theta whose tangent is this
quantity by reading off the cosecant and
secant from this triangle.
So the co-secant remember is a one over
sine which means it's hypotenuse divided
by the opposite side length and that's
what I got here, length by hypotenuse
divided by length of opposite side and
secant, remember is one over cosine so
that means it's the side length by
hypotenuse divided by this length, which
is just one.
Now once I know cosecant and secant in
terms of this quantity here, I can then
evaluate L at that quantity.
So here we go, this is what I get when I
plug in these quantities for coseekit, and
seekit.
Get.
Now this looks terrible, right?
Well, I can simplify this a little bit.
What do you notice here?
I've got a dividing by the cubert of B in
the denominator, which is as good as
putting a B to the thirds power up here.
And here I'm going to separate out this B
to B a B to the two thirds times B to the
one third and move that inside the square
root.
The upshot is that this thing is in fact
the same as this thing here.
And now this begins to be a little bit
more symmetric, right?
I took the a and the cube root of a and
that gave me the a to the 2 3rds and by
putting a b to the 2 3rds inside here, I
made this look a lot, a lot nicer.
Okay.
Now I've got a common factor of the square
root of b to the 2 3rds plus a to the 2
3rds here and I can collect that out.
So now I've got a to the 2 3rds plus b to
the 2 3rds times the square root of b to
the 2 3rds plus a to the 2 3rds.
Now this is the same thing, here.
It's this thing times the square root that
thing.
And another way of saying that is just
this quantity to the 3/2 power, right?
This is this quantity to the first power,
times the 1/2 power, and that gives me the
3/2 power, here.
So, at the critical point, l of theta is
equal to this quantity.
Now with a bit more work, we can show that
this quantity is in fact the minimum value
of l.
We can summarize our answer.
So given a hallway of width A meeting a
hallway of width B at some corner, what
we've shown is that a stick of length at
most, A to the two thirds plus B to the
two thirds, that quantity to the three
halves power can be navigated around that
corner.
And we can see this playing out in action.
I made a model that you can download Well
here is an example.
I've got a hallway of width 64
millimeters, that's a pretty small
hallway, right?
And I've got another hallway here, 27
millimeters.
And the calculation that we've done tells
us that a stick of length 125 millimeters
will just be able to turnaround, this
hallway.
So let's see.
Here's the stick and I start the stick
moving down the hallway, right?
And I start trying to rotate it around and
eventually I'm going to bump up against
the corner , but I'll just be able to make
it around the corner and then I'll be able
to move it out of the hallway.
Perhaps you're wondering where those
numbers came from.
Well here's where I got these numbers
from, right?
64 and 27, these are the widths of my
hallways and this is the formula that
tells us the largest stick that can pass
through a hallway of these two widths.
Now I chose 64 and 27 for, maybe two
different reasons.
One thing is that these are both perfect
cubes.
So, 64 to the one third power is 4 and 27
to the one third power is 3.
The other thing that's nice about these
numbers is that 4 squared plus 3 squared
is 5 squared, right?
These are Pythagorean triple.
And 5 squared to the 1/2 is just 5.
So it's 5 cubed which is 125.