Shortest distance between 2 points 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. 
 A: 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.
A: 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.
