Hi everyone. This question comes from the 2001 senior competition and only 4% of
students got this one right. So we need to find the minimum value of that long
algebraic expression which I'm not going to read out, so pause the video here and
see if you can figure it out. Alright so remember to like and subscribe if you
haven't already now onto my solution. So I'm gonna go into quite a lot of detail
about my thought process as I went through this question because I think
it's like, it's really obvious once you see the solution but it's not obvious
how to actually get there. So what I noticed was that this looks a lot like
Pythagoras' theorem, like if you imagine a right-angled triangle with x there and
one then that hypotenuse is square root x squared plus 1 so this might be
helpful because if we can interpret that expression geometrically then that might
actually help us to work out the minimum value rather than having to resort to
calculus or something which is gonna be really complicated for this question. Ok
so that's that one. For this one it would look like, you know, you've got y
minus x there, and then that's 2 squared so 2 would be the other side and then
that length is that one. So I think my original thought with this was that
maybe we'd like arrange all these triangles to make a quadrilateral. I
don't know why I thought that or like how that would be useful but that was
kind of my initial idea. I think it's really important with these to just sort
of like do something, even if you don't really have a clear idea of where you're
going because, you know, as you draw stuff and move things around you might be able
to figure it out. Like you don't have to have a clear plan when you start out. So
I think the next thing I did was, I thought about this y minus x and
what that might mean. So if we've got a length y
and we subtract a length then what we have left is the y minus x. So the next
thing I did was to draw out like if that whole length is y, from there to there, and
then we've got a small length x inside that then this bit here will be y minus
x. So we might draw it kind of like... there's the x, there's the 1, and then that's
that length and then over here you'll have the y minus x and the 2 up there.
And then there's... if we had like that whole length is z, and from
there to there is y, then this bit will be z minus y, and so on, so I kind
of... I drew that up and then that wasn't really helpful because what we want is
to find the sum of all of these so really what I want is for this one to
start where this one ends because then like the sum of these two will be sort
of the distance traveling along there. So then what I did was I decided to shift
this triangle up so just shift this up to there, so let me just get rid of that.
So instead of starting it on this line, let's put it there, so now we've got the
2 there, that distance is y minus x, and now this is the square root of y minus x
squared plus 2 squared. Okay so let's put on the z minus y. Let me just shift that
to get some space. So for this one we again wanna start that out here so that we
can add all these lengths together, like now this sum is gonna be the distance as
you travel along this thing.
So that one was 1. Now you can also think of these... instead of like drawing
out you know that whole thing is y and so on we could think of it in terms of
coordinates, so like that point's y, that one's x, that one's z, and then if we
want 10 minus z then that's gonna be 10. So that's probably a better way of
thinking of it. And like that point is 0. Okay and then that's 3 squared
so for this one we want to have a triangle with a height of 3, where that's 10
minus z. Okay and then we want to find the minimum value of that total
distance as you travel along here. So we can pick x, y, and z to be whatever we want in
order to minimise that distance, but no matter what values of x, y, and z you choose,
you're going to be traveling from here to there and that point is always gonna
be at the same place, because that's at ten and that height is set by these
heights, so the shortest distance from there to there is just gonna be a
straight line like that. Now it doesn't really matter what x, y, z are; we weren't
asked to find them, but you could find them if you wanted just by doing similar
triangles, like if you add up all these heights we've got 1, 2, 1, and 3 so that
total height is 7, and that distance is 10 so all these little triangles... if we
actually move x so that... that height still needs to be 1... x should really be
there for it to stay on that straight line.
And then, you know, if you wanted to work out what x was you could just do similar
triangles, like that side to that side, so x to 1 has to be in the same ratio
as 10 to 7. But it doesn't really matter because all that we want to work
out is that minimum value of that, so the distance from there to there, which we can
just do using pythagoras, so square root of that side squared, 10 squared, plus
that side squared, which is 7, so that makes root 149, and then we can't simplify that
at all so that is the answer to that question. So I had another look at the
question and tried out doing it with calculus, and you know it's actually not
THAT bad. I knew from the start that they really didn't intend you to use calculus
to do it, because I don't think you need calculus for any of the questions in the
Australian Mathematics Competition, because they wouldn't want to
disadvantage the students who haven't learned it yet.
But okay, so what I did was I started out doing the partial derivative of that
thing with respect to x. Now I know most of the high school students won't have
done partial derivatives. It just means that we sort of consider the y and the z
to be constants and just differentiate this with respect to x, and it's just
like this funny d notation. Okay so I did that. You want to set that derivative to
be zero because that's what's gonna minimise that whole thing. Yadda yadda yadda,
lots of algebra. I got down to this point. Now I knew from the start that we'd want
each of these things to be positive so I made that assumption there. The reason
for that is that like we want all this to be as small as possible,
and so we want like this to be small, and this to be small, and this to be small, and
so on, now this is kind of like the distance between 10 and z and this is
like the distance between z and y and y and x and x and 0, so if you imagine on a
number line like here's 0, here's 10, and we want to place x, y, and z so that the
distance from zero to x is minimised and x to y is minimised and so on
then you'd want x, y, and z to be like lying between 0 and 10 in that order. Actually
the first thing I tried, even before I did, you know, the triangles and stuff, was
that I thought, you know, maybe these would be equally spaced so x would be 2.5, y
would be 5, and z would be 7.5, so I was pretty comfortable making that
assumption and then that makes it a lot easier because then we only need to
consider the plus there and not the minus. So that gives you y = 3x, and
then you do the derivative with respect to y as well, and make a similar
assumption, and you get the second equation. And then with respect to z
and so on, and then you've got these three equations and you can solve those
simultaneously and that gives you these values and then you can plug those in
and get to that. So it was obviously a lot more messy than the geometric
solution, but you CAN do it with calculus. Oh and I should explain why it's not the
case that these are just equally spaced along here. So that's because all these
numbers are different and that effectively gives different weightings
for the need for each of these intervals to be as small as possible. So it
actually ends up with like x being quite low down there and then there's y and
z there because that number is bigger, that makes it so that like this interval,
the 10 minus z, can be bigger, kind of. That doesn't need to make intuitive
sense to you, it's just... that's why we have to do all the calculus really.
All right so leave me a comment below if you still have any questions and give
this video a thumbs up if you want me to make more like this, and stay tuned for
the next one. I haven't picked out a question yet but I'm sure it's gonna be
a fun one.