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Ok, this problem might be trivial, but i'm a bit stuck.

A string with length 3m is shaped to an "L". Find the shortest distance between the 2 endpoints.

So what i've done is to draw the figure, and i can see that pythagoras is key here. But how do i find the shortest distance?

Thanks.

kirkegaard
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    It depends on the font used for the L. – Regret Feb 16 '15 at 09:33
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    Haha, i should put two lines under that answer and turn in – kirkegaard Feb 16 '15 at 09:34
  • @Regret: more to the point, the font used for the L is the variable we're qualifying over when we say "the shortest". – Steve Jessop Feb 16 '15 at 12:47
  • @Steve: In many fonts, $\cal L$ is not displayed as two line segments perpendicular to each other. – Regret Feb 16 '15 at 12:51
  • @Regret: then we have a choice either to include all of them in our consideration, or else get the problem finished before the deadline ;-) – Steve Jessop Feb 16 '15 at 12:52
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    @Steve: We could also trivialize the problem by defining a font where $\cal L$ is a circle. – Regret Feb 16 '15 at 12:54
  • @Steve: Apologies if I seem rude, I find joy in what may be considered pedantry. – Regret Feb 16 '15 at 13:01
  • @Regret: not at all rude. I suspect the professor will mark it wrong if you come back and say "as far as the fonts on my PC are concerned, the distance is minimised for the Wingdings font", or whatever it might be. But he might also change the question in future, to clarify that the English phrase "shaped to an L" means something other than "shaped like any glyph used to represent the byte with value 76, in any font you can create". – Steve Jessop Feb 16 '15 at 13:07

2 Answers2

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Let $d$ be the distance between the two endpoints.

Let say that the angle of the $L$ shape is at distance $x$ from the first extremity. I suppose that the angle is $\dfrac{\pi}{2}$.

Then you have, with Pythagoras helping a bit:

$d=\sqrt{x^2+(3-x)^2}=f(x)$

$f'(x)=\dfrac{4x-6}{\sqrt{x^2+(3-x)^2}}$ which has an extremum for $x=\dfrac 32$. $f'(x)$ is negative for $x<\dfrac 32$ so the extremum is a minimum.

EDIT: typo modified, the angle is indeed $\dfrac{\pi}{2}$ and not $\pi$. Thanks for pointing it out.

Martigan
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From a purely analytic angle, we can let $l$ and $w$ be the length and width of the L, giving the equations $$l+w=3$$ and $$l^2+w^2=D^2$$ where $D$ is the distance between the endpoints. Substituting, we have $$w=3-l$$ and so $$l^2+(3-l)^2=D$$ $$2l^2-6l+9=D.$$ Since we want to minimize $D$, and we have a positive parabola, we find the vertex ${6 \over 2\cdot 2}=1.5$. Thus we have $l=w=1.5$ and so $D=1.5\sqrt 2\approx 2.12$.

Alternatively, we can note that the diagonal will always be smallest in a square of any rectangle, so $l=w=1.5$ and so on. Depends on your tastes.

Avi
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    "we can note that the diagonal will always be smallest in a square of any rectangle" -- the proof of which can in any case be the same as what you did to answer the problem directly, but with a named value in place of $3$. – Steve Jessop Feb 16 '15 at 12:50
  • @SteveJessop Yes, this is essentially the rigorous proof of that fact; my point was that if we view the problem as asking, "What rectangle has the shortest diagonal relative to its perimeter?" we can intuitively see the answer without calculation (though also without rigor). – Avi Feb 16 '15 at 13:08