PROFESSOR: Welcome
back to recitation. In this video we'd like to do
another optimization problem. This one's a little bit harder
than the distance problem. So the question
is the following: consider triangles formed by
lines passing through the point x-- (8, 4), sorry, the
x-axis and the y-axis. Find the dimensions
that minimize area. So what does this
fist sentence mean? It really means use
this point to draw a line through this point--
I'll give you an example, it's kind of a wiggly line, but
hopefully it looks like a line to you-- and it makes a triangle
with this line, the x-axis, and the y-axis. We can certainly calculate
the area of that triangle. So the problem is
asking you to find the dimensions of
the triangle that minimize the area
with the constraint that the line, the hypotenuse
goes through the point (8, 4). I'm going to give you a
couple minutes to work on it. Why don't you pause video here
and then when you're ready, restart the video,
I'll come back, and I'll help you
solve the problem. Welcome back. So again, we're doing
an optimization problem. And we want to optimize--
because it says minimize area, we know the optimizing
equation is area. So let's be very clear. Always, when you're
doing these problems, you have, again, as
we've said previously, you have a constraint
equation and you have an optimizing equation. The optimizing equation now,
we've already said, is area. And area, the easiest way
to write area in this form is-- notice that this distance,
we could write it as base times height or we could
write it as x times y-- so the base here is x
and the height here is y. So the area of the triangle
is 1/2 base times height. So the area is 1/2 x times y. That's the thing we
want to optimize. The problem is that we know when
we're doing these optimization problems we want to take
a derivative of area with respect to a variable, but
right now we have two variables and so that's where the
constraint equation comes in. So now we have to figure out how
we're going to use a constraint equation here. The constraint is that it has to
go through this point, (8, 4). So what does our line
have to look like? Well, our line has to look like,
ultimately-- let's do, maybe, the point-slope form. y is equal-- or sorry. I said point-slope form. y minus 4 is equal
to m times x minus 8. Right? Notice I couldn't
pick what m was. Because the m completely
determines the line. So hopefully that make
sense, that you can see that. Now, in fact, let's look at
how this problem will work. The m is going to
determine this point and it's going to
determine this point. If you can't see that,
well, let's look back here. This point is when y equals 0. Right? So I can put in y equals 0
and I get x in terms of m. If I come back over here
and look at this point, this is when x equals 0. So if I put in 0 for x, I
can find y in terms of m. So these two values, the
x-value and the y-value, completely determined on
the slope of this line. That hopefully makes
sense just even if you look at the geometric picture. When I turn about this point
at (8, 4) these values change. So the x and y
values are completely determined by the
slope of the line. In fact, the area,
then, is completely determined by the
slope of the line. So what we're going
to do is we're going to use the
constraint equation to find x and y values,
all in terms of the slope. So let's do that. I said when y is 0,
what do we get for x? We get negative 4 over
m plus 8 is equal to x. Let me double check my math so
I don't have to re-shoot this. When y is 0 I divide
by m, I add 8, I get x. So that is the x-value I'm
interested in down here. When x is 0-- let's see
what I get-- when x is 0 I get negative 8m plus 4 is y. Right? x is 0, negative 8m plus 4. So now what I'm going to
do is plug these two things into the area equation. Area is now equal
to 1/2 of x times y. So 1/2 of 8 minus
4 over m times-- you know what I'm going to do? I'm going to take this 1/2
and kill off terms in there so I don't have to
worry about it anymore-- negative 4m plus 2. So this is x and
this is half of y. So just to make it simpler I'm
not carrying through the 1/2-- I'm killing off
half of the things, dividing every term in y by 2. And again, what are
we trying to do? We're trying to optimize. So now we want to take
the derivative of area with respect to the slope. So this is-- maybe to simplify
first, let's multiply through. So this is just a little
bit of algebra really quick. 8 times 4 is 32, so I
get negative 32m plus 16. And then here, negative
times negative is a positive. 4 times 4 is 16. m divided by m, I just get 16. And then here I get negative 8m. So I had to do a little
bit of algebra first, but this is much easier
to take a derivative and not make mistakes
than this one. Because you'd have a
product rule and then you'd still have to multiply. So we might as well
multiply out first. So now let me just take
the derivative of this. And again, I'm taking the
derivative with respect to m. So here I just get
negative 32, 0, 0, and then what's the
derivative of-- this is a minus 8m-- well, the
derivative of 1 over m, if you remember, is
negative 1 over m squared. I have another negative
here, so this is going to be plus 8 over m squared. Right? Optimizing, we want to set
the derivative equal to 0. So if I set the
derivative equal to 0 and solve I get 32 m squared
equals 8, or m squared is equal to 8 over 32, which
is 1/4, or m is equal to 1/2. Or I should say,
plus or minus 1/2. We need to be aware. I would run into problems
if I didn't put the minus. So solving this problem, I see
that-- again, what did I do? I set area prime equal to 0,
move the 32 over, multiply by m squared, do
some algebra, and I get m is equal to
plus or minus 1/2. And now we need to see which
of these things make sense and then we just need to
think about what happens as m goes to its extreme values. So let's come back and
look at the picture and from there we can probably
tell which of these answers we need. So it's m equals 1/2 or
m equals minus 1/2 that we want to know which
of these do we need. So I'm going to use some
different colored chalk to draw what's happening here. Notice the slope of
this line is negative. Right? If I were going to do a
positive sloping line, which would be the case where
m is equal to 1/2, I would get something that's
headed in this direction. And notice that that's not going
to make a triangle with the x- and y-axis. And so immediately m equals
1/2 isn't even in this problem, isn't allowed to work. OK, now where did it come from? It came because somewhere
I was multiplying m by itself, which maybe isn't
actually in the original part. I was introducing a new
thing happening, there, so I'm not going to get
into it too much because we can immediately see
that we don't have to worry about m equals 1/2. m equals minus 1/2 looks good. That's sloping in
this direction. And in fact, that would
give us a nice triangle. The extreme values in
this case are obviously when m is sloping all the
way up to being vertical, or when m is sloping
to being horizontal. And in both of those cases you
notice that the area is getting arbitrarily large, it's headed
towards infinity in both cases. So I don't need to worry about
looking at the extreme values. There aren't end points
really in this case. But the extreme
values, they're both going to positive
infinity, the areas. Which convinces me
even more that where m is equal to minus 1/2
is going to be a minimum. You could also take
the second derivative and run the second derivative
test, but even geometrically, we can see in the picture
that at m equals negative 1/2 we actually get a
negative sign for the-- or, sorry-- a
minimizer for the area. And now the question asks
to find the dimensions. How do I go back and
find the dimensions? I'm not going to do any
more on this problem, but you can do it
to finish it off. Finding the dimensions,
I know what m is. I also know what
x is in terms of m and what y is in terms of m. So I just evaluate x at the
m and evaluate y at that m. That gives me the dimensions
that will complete the problem. But I think I'll stop there.
