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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 point
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x-- (8, 4), sorry, 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
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a line through this point--
I'll give you an example,
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it's kind of a wiggly line, but
hopefully it looks like a line
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to you-- and it makes a triangle
with this line, the x-axis,
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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
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the dimensions of
the triangle that
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minimize the area
with the constraint
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that 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,
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and I'll help you
solve 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--
because it says minimize area,
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we know the optimizing
equation is area.
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So let's be very clear.
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Always, when you're
doing these problems,
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you have, again, as
we've said previously,
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you have a constraint
equation and you
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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
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is-- 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--
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so the base here 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 optimization
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problems we want to take
a derivative of area
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with respect to a variable, but
right now we have two variables
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and so that's where the
constraint equation comes in.
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So now we have to figure out how
we're going to use a constraint
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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,
ultimately-- let's do, maybe,
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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
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and it's going to
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,
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this is 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 determined on
the slope of this line.
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That hopefully makes
sense just even if you
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look at the 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
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determined by the
slope of the line.
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In fact, the area,
then, is completely
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determined by the
slope of the line.
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So what we're going
to do is we're
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going to use the
constraint equation
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to find x and y values,
all in 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, 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-- let's see
what I get-- when x is 0
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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
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into the 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
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so I 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 half 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,
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dividing every 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
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with 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,
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but this is much easier
to take a derivative
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and not make mistakes
than this one.
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Because you'd have a
product rule and then
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you'd still have 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, 0,
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0, and then what's the
derivative of-- this
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is a minus 8m-- well, the
derivative of 1 over m,
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if you remember, is
negative 1 over m squared.
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I have another negative
here, so this is going
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to be plus 8 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
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and solve I get 32 m squared
equals 8, or m squared
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is equal to 8 over 32, which
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
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by m squared, do
some algebra, and I
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get m is equal to
plus or minus 1/2.
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And now we need to see which
of these things make sense
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and then we just need to
think about what happens as m
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goes to its extreme values.
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So let's come back and
look at the picture
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and from there we can probably
tell which of these answers
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we need.
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So it's m equals 1/2 or
m equals minus 1/2 that
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we want to know which
of these do we need.
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So I'm going to use some
different colored chalk
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to 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,
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I would get something that's
headed in this direction.
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And notice that that's not going
to make a triangle with the x-
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and y-axis.
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And so immediately m equals
1/2 isn't even in this problem,
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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
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by itself, which maybe isn't
actually in the original part.
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I was introducing a new
thing happening, there,
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so I'm not going to get
into it too much because we
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can immediately see
that we don't have
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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
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when m is sloping all the
way up to being vertical,
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or when m is sloping
to being horizontal.
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And in both of those cases you
notice that the area is getting
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arbitrarily large, it's headed
towards infinity 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
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going to positive
infinity, the areas.
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Which convinces me
even more that where
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m is equal to minus 1/2
is going to be a minimum.
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You could also take
the second derivative
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and run the second derivative
test, but even geometrically,
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we can see in the picture
that at m equals negative 1/2
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we actually get a
negative sign for the--
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or, sorry-- a
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,
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but you can 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
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and what y 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 complete the problem.
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But I think I'll stop there.