Hey so I'm having trouble with this whole question. Can you show me proving? Thanks heaps.
Ergh, this one is nasty
Okay, so let's consider the distance \(L\) in two parts: The length of rope PM and the distance PX.
Let's start with PX. That's just pythagoras on the triangle on the left:
Now, PM is the length of the rope in total, minus the length of rope AP:
Now we can find AP using pythagoras theorem, noting that \(AX=1-x\):
^2+r^2-x^2}\\=\sqrt{1-2x+r^2})
So, \(PM=2-\sqrt{1-2x+r^2}\).
Putting all that together:
And that is the proof. From here it is just normal (albeit tedious) differentiation work.
Then you just use the factorisation given to solve for the numerator equal to zero:
(2x^2-r^2x-r^2)=0)
So \(x=1\) is one obvious candidate; but it's actually incorrect, the rope would just be hanging that way, it violates our conditions!
Note that here I deviate from the answer above, because the factorisation above was incorrect. For the right bracket, use the quadratic formula:
Note that we don't care about the negative answer here, since x>0
Now this is the value we want, and
we do need to prove that it is a maximum. You could do this by differentiating again (10/10 disgusting) or more nicely, just substitution into the first derivative. Tedious algebra, but the usual technique!
(Ps - Great answer kb123! Only uploaded my Part 1 and 2 because I had it ready
)
PS - This is far more difficult than anything you will get in 2U these days, this is from a 2003 HSC Paper 
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