WEBVTT

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PROFESSOR: Welcome back
to recitation.

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In this video we'd like to do
another optimization problem.

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This one's a little bit harder
than the distance problem.

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So the question is
the following--

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consider triangles formed by
lines passing through the

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point x, h, 4-- sorry--

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the x-axis and the y-axis.

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Find the dimensions that
minimize area.

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So what does this fist
sentence mean?

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It really means use this
point to draw a line

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through this point--

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I'll give you an example, it's
kind of a wiggly line, but

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hopefully it looks like
a line to you--

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and it makes a triangle
with this line, the

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x-axis, and the y-axis.

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We can certainly calculate the
area of that triangle.

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So the problem is asking you to
find the dimensions of the

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triangle that minimize the area
with the constraint that

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the line, the hypotenuse goes
through the point 8, 4.

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I'm going to give you a couple
minutes to work on it.

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Why don't you pause video here
and then when you're ready,

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restart the video, I'll come
back, and I'll help you solve

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the problem.

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Welcome back.

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So again, we're doing an
optimization problem.

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And we want to optimize--

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because it says minimize area,
we know the optimizing

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equation is area.

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So let's be very clear.

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Always, when you're doing these
problems, you have,

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again, as we've said previously,
you have a

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constraint equation and you have
an optimizing equation.

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The optimizing equation now,
we've already said, is area.

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And area, the easiest way to
write area in this form is--

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notice that this distance, we
could write it as base times

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height or we could write it as
x times y-- so the base here

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is x and the height here is y.

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So the area of the triangle
is 1/2 base times height.

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So the area is 1/2 x times y.

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That's the thing we
want to optimize.

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The problem is that we know
when we're doing these

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optimization problems we want
to take a derivative of area

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with respect to a variable,
but right now we have two

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variables and so that's
where the constraints

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equation comes in.

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So now we have to figure out
how we're going to use a

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constraint equation here.

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The constraint is that it has to
go through this point 8, 4.

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So what does our line
have to look like?

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Well, our line has to look
like ultimatlely--

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let's do, maybe, the
point-slope form--

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y is equal, or sorry.

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I said point-slope form.

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y minus 4 is equal to
m times x minus 8.

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Right?

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Notice I couldn't
pick what m was.

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Because the m completely
determines the line.

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So hopefully that make sense,
that you can see that.

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Now, in fact, let's look at how
this problem will work.

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The m is going to determine this
point and it's going to

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determine this point.

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If you can't see that, well,
let's look back here.

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This point is when y equals 0.

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Right?

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So I can put in y equals 0 and
I get x in terms of m.

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If I come back over here and
look at this point, this is

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when x equals 0.

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So if I put in 0 for x, I can
find y in terms of m.

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So these two values, the x
value and the y value,

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completely determine on the
slope of this line.

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That hopefully makes sense just
even if you look at the

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geometric picture.

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When I turn about this point at
8, 4 these values change.

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So the x and y values are
completely determined by the

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slope of the line.

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In fact, the area, then, is
completely determined by the

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slope of the line.

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So what we're going to do is
we're going to use the

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constraint equation to find
x and y values, all in

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terms of the slope.

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So let's do that.

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I said when y is 0, what
do we get for x?

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We get negative 4 over m
plus 8 is equal to x.

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Let me double check my math so I
don't have to re-shoot this.

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When y is 0 I divide by m.

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I add 8, I get x.

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So that is the x value I'm
interested in down here.

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When x is 0--

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let's see what I get--

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when x is 0 I get negative
8m plus 4 is y.

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Right?

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x is 0, negative 8m plus 4.

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So now what I'm going to do is
plug these two things into the

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area equation.

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Area is now equal to
1/2 of x times y.

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So 1/2 of 8 minus
4 over m times--

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you know what I'm going to do?

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I'm going to take this 1/2 and
kill off terms in there so I

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don't have to worry
about it anymore--

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negative 4m plus 2.

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So this is x and this
is 1/2 of y.

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So just to make it simpler I'm
not carrying through the 1/2--

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I'm killing off half of the
things, dividing every

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term in y by 2.

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And again, what are
we trying to do?

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We're trying to optimize.

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So now we want to take the
derivative of area with

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respect to the slope.

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So this is, maybe to simplify
first, let's multiply through.

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So this is just a little bit
of algebra really quick.

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8 times 4 is 32, so I get
negative 32m plus 16.

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And then here, negative times
negative is a positive.

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4 times 4 is 16.

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m divided by m, I just get 16.

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And then here I get
negative 8m.

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So I had to do a little bit of
algebra first, but this is

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much easier to take a derivative
and not make

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mistakes than this one.

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Because you'd have a product
rule and then you'd still have

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to multiply.

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So we might as well multiply
out first.

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So now let me just take the
derivative of this.

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And again, I'm taking the
derivative with respect to m.

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So here I just get
negative 32.

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0, 0.

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And then what's the
derivative of--

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this is a minus 8m--

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well, the derivative of 1 over
m, if you remember, is

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negative 1 over m squared.

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I have another negative here, so
this is going to be plus 8

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over m squared.

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Right?

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Optimizing, we want to set the
derivative equal to 0.

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So if I set the derivative equal
to 0 and solve I get 32m

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squared equals 8, or m squared
is equal to 8 over 32, which

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is 1/4, or m is equal to 1/2.

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Or I should say, plus
or minus 1/2.

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We need to be aware.

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I would run into problems if
I didn't put the minus.

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So solving this problem, I see
that-- again, what did I do?

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I set area prime equal to 0,
move the 32 over, multiply by

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m squared, do some algebra, and
I get m is equal to plus

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or minus 1/2.

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And now we need to see which of
these things make sense and

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then we just need to think about
what happens as m goes

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to its extreme values.

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So let's come back and look at
the picture and from there we

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can probably tell which of
these answers we need.

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So it's m equals 1/2 or m equals
minus 1/2 that we want

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to know which of these
do we need.

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So I'm going to use some
different colored chalk to

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draw what's happening here.

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Notice the slope of this
line is negative.

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Right?

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If I were going to do a positive
sloping line, which

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would be the case where m is
equal to 1/2, I would get

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something that's headed
in this direction.

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And notice that that's not going
to make a triangle with

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the x- and y-axis.

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And so immediately m equals
1/2 isn't even in this

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problem, isn't allowed
to work.

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OK, now where did
it come from?

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It came because somewhere I was
multiplying m by itself,

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which maybe isn't actually
in the original part.

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I was introducing a new thing
happening, there, so I'm not

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going to get into it too much
because we can immediately see

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that we don't have to worry
about m equals 1/2.

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m equals minus 1/2 looks good.

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That's sloping in
this direction.

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And in fact, that would give
us a nice triangle.

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The extreme values in this case
are obviously when m is

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sloping all the way up to being
vertical, or when m is

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sloping to being horizontal.

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And in both of those cases you
notice that the area is

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getting arbitrarily large, it's
headed towards infinity

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in both cases.

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So I don't need to worry about
looking at the extreme values.

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There aren't end points
really in this case.

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But the extreme values, they're
both going to positive

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infinity, the areas.

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Which convinces me even more
that where m is equal to minus

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1/2 is going to be a minimum.

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You could also take the second
derivative and run the second

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derivative test, but even
geometrically, we can see in

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the picture that at m equals
negative 1/2 we actually get a

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negative sign for the--
or, sorry-- a

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minimizer for the area.

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And now the question asks
to find the dimensions.

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How do I go back and find
the dimensions?

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I'm not going to do any more on
this problem, but you can

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do it to finish it off.

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Finding the dimensions,
I know what m is.

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I also know what x is in
terms of m and what y

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is in terms of m.

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So I just evaluate x at the m
and evaluate y at that m.

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That gives me the dimensions
that will

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complete the problem.

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But I think I'll stop there.