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CHRISTINE BREINER: Welcome
back to recitation.

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Today we're going to work
on an optimization problem.

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So the question I
want us to answer

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is, what point on the curve y
equals square root of x plus 4

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comes closest to the origin?

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I've drawn a sketch
of this curve.

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The scale in this
direction-- each hash mark

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is one unit in the x
direction, each hash mark here

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is one unit in the y direction.

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Just want to point
out two easy places

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to figure out the
distance to the origin.

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Over here, where the curve
starts at negative 4, 0,

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the distance to the
origin is 4 units.

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And here at (0, 2) the distance
to the origin is two units.

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It's probably, we could
safely say, further away here.

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So we're anticipating
that somewhere

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along the curve
in this region is

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where we should find our place
that's closest to the origin.

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The only reason I
point that out is that,

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when you're doing these problems
on your own you should always

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try and anticipate roughly
where things should happen,

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in what kind of region, so that
you don't-- you don't start

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thinking, if you do something
wrong and you get x equals 100

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and then you come back
and look at the curve,

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you realize right away, well,
that doesn't make any sense.

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So we want to always be thinking
as we're solving the problems,

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does my answer make sense?

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So I'm actually going to
give you a little bit of time

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to work on this yourself
and then I'll come back

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and I'll work on it as well.

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Welcome back.

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Hopefully you were able to get
pretty far into this problem.

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And so I will start
working on it now.

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So again, the
question is that we

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want to optimize-- in this
case, minimize-- the distance

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to the origin from this curve.

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And so what we're really trying
to do is we have a constraint,

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the constraint is we
have to be on the curve,

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and then we also have something
we're trying to minimize.

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And the thing we're trying
to minimize is distance.

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And so we have to make sure that
we understand the two equations

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that we need-- the optimization,
or the constraint equation,

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and the optimizing equation.

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So to optimize we need to
know how to measure distance

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in two-dimensional space.

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And one point I want
to make is that if you

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want to optimize distance
you might as well

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optimize the square of distance
because it's much easier.

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So let me justify that
briefly and then we'll go on.

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So I want to optimize the
distance squared to the origin.

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It's, well distance, you
remember, first in general,

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between two points
(x, y) and (a, b)

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is something in this form.

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Distance squared
is the difference

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between the x-value
squared plus the difference

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between the y-value squared.

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This is, should remind you
of the Pythagorean theorem,

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ultimately.

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So in this case, in our
case, distance to the origin

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is x squared plus y squared.

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The distance squared is
x squared plus y squared.

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I just told you that instead
of trying to optimize distance,

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we can optimize
distance squared.

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Why is that?

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Well, remember that
when you optimize,

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what you're looking
for is a place where

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the derivative of the function
of interest is equal to 0.

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So what I want to
point out is that when

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you take the
derivative of distance

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squared and find
where that's 0, it's

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going to be the
same as the place

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where the derivative of
distance is equal to 0.

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So let's notice that.

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So this is a little
sidebar justification.

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Notice d squared prime
is equal to 2d d prime.

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Where did that come from?

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That's this is implicit
differentiation

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with respect to x and
this is the chain rule.

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So if I want d
prime to equal 0, I

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can also find where d
squared prime equals 0.

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I'm assuming-- notice
the distance is never

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at the origin-- so
distance is never 0.

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So I don't have to
worry about that.

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So that's a small
sidebar, but just

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to justify why we can do that.

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Now let's come back into
the problem at hand.

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What is our optimization
problem, equation

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that we want to minimize?

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We want to minimize this
equation with respect

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to a certain constraint.

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What's the constraint?

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The constraint is what y is.

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y depends on x.

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And so when I solve
these problems

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I'm going to have to
substitute in my constraint.

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So y squared is the square root
of x plus 4 quantity squared,

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so I just get x plus 4.

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So now I have my
optimization equation.

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How do I find a
minimum or a maximum?

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I take the derivative
and set it equal to 0.

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So let me come give
myself a little more room

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and do that over here.

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So d squared prime, now I get
derivative of x squared is 2x.

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The derivative of x is 1,
and the derivative of 4 is 0.

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This will be optimized
when this is equal to 0.

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So 0 equals 2x plus 1.

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So x is equal to minus 1/2.

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Does this pass, as we would
say maybe, the smell test?

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Does it smell OK to us?

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The answer will be yes.

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Because remember, we said
somewhere in this x region

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is where we expect that we
will have a distance closest,

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point closest to the origin.

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And so we're right
here on the x value.

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Now we have to find what the y
value is to finish the problem.

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But this is not, so
far, very surprising.

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It seems like maybe
the right thing.

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Now we have x.

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So now how do we find y?

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Well, we know what y is.

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y is equal to the
square root of x plus 4,

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so it's equal to the square
root of negative 1/2 plus 4,

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which simplified is 3 and
1/2, which I think is 7/2.

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So the point is negative 1/2
comma square root of 7/2.

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And then you just want to
double check and make sure,

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did I ask for the distance
or did I ask for the point?

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And right now we have the
point, so let's come over

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and make sure what
point on the curve

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comes closest to the origin.

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So now we know that we've
answered the correct question.

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So again, it was a
maximize-- sorry,

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it was a minimizing problem.

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It was an optimization
problem where

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we wanted to minimize distance.

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We had a constraint equation.

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We had the thing we
wanted to minimize.

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And then we took the
derivative of the minimizer,

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set it-- of the optimizing
equation, set it equal to 0,

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solved for x, and then found the
answer to the specific question

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by then finding the y-value.

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And I think I'll stop there.