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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 is, what point on the

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curve y equals square root
of x plus 4 comes

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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--

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

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mark here is one unit
in the y direction.

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Just want to point out two easy
places to figure out the

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distance to the origin.

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

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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 along the curve in

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this region is where we should
find our place that's closest

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to the origin.

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The only reason I point that out
is that, when you're doing

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these problems on your own
you should always try and

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anticipate roughly where things
should happen in what

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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, you realize

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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

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problems, does my answer
make sense?

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

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

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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 want to optimize--

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in this case, minimize--

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the distance to the origin
from this curve.

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

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constraint , the constraint is
we have to be on the curve,

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and then we also have something

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we're trying to minimize.

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

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

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equations that we need--

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the optimization, or constraint
equation and the

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optimizing equation.

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So to optimize we need to know
how to measure distance in

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two-dimensional space.

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And one point I want to make
is that if you want to

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optimize distance you might as
well optimize the square of

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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

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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
is something in this form.

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Distance squared is the
difference between the x value

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squared plus the difference
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 is x

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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

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distance, we can optimize
distance squared.

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Why is that?

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Well, remember that when you
optimize, what you're looking

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for is a place where the
derivative of the function of

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interest is equal to 0.

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So what I want to point out
is that when you take the

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derivative of distance squared
and find where that's 0, it's

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going to be the same as the
place where the derivative of

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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 g 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 with respect

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to x and this is
the chain rule.

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So if I want d prime to equal
0, I can also find where d

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squared prime equals 0.

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I'm assuming-- notice the
distance is never at the

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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 to justify

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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 that we want

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to minimize?

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We want to minimize this
equation with respect to a

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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 I'm going to have to

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substitute in my constraint.

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

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squared 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 and

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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 is

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

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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

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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, so it's equal to

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the square root of negative 1/2
plus 4, which simplified

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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 and

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make sure what point
on the curve comes

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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--

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sorry, it was a minimizing
problem.

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It was an optimization
problem where we

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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, or of optimizing
equation, set it equal to 0,

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

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question by then finding
the y value.

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And I think I'll stop there.

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