Math Is Fun Forum

Hi c6_viyen_1995,

Graph this function y = (1+1/x)^(x) here http://www.mathsisfun.com/data/grapher-equation.html.
It is always increasing in (0,∞). You can prove this using calculus.
Post again if you have any problem.

Hi Brent,

Check this page when you have some free time.
http://www.mathisfunforum.com/viewtopic.php?id=15139
There are some good points that will help you graph functions.

Hi bobbym,
Thank you. Nice problem.

Hi gar, Yes that is how i did it too!

Second try-

Hi gAr,

Consider three ants at the end of an equilateral triangle of side length = s moving towards each other at constant speed = v.

As the ants move towards each other, points joining the three ants will be an equilateral triangle of a smaller side.(by symmetry)
As time passes the side of this equilateral triangle decreases.

If you choose any of the ant as your frame of reference, the component of velocity of the other ants towards the fixed (frame) ant
will be v + v/2. where v is the speed of the ant. [because velocity of A wrt. B is equal to  velocity of A wrt. F +  velocity of F wrt. B].
This means that the fixed ant will see the others coming closer to it at speed = 3/2v.

And in the ant frame the other ants travel towards the fixed one in a straight line. Therefore distance traveled(in ant frame) = s.

Time for which the ants move is = T = dist/speed = s/(3v/2).

In the ground frame they move with speed v for a time = T. Therefore distance traveled by the ants = vT = (2/3)s

Oh! I also forgot to hide the ans. That was my first post using latex!

Hi !!!!

I am back!

hi bobby,

Hi Sameer,
Bobby was just teaching me a new optimisation tool called the Method of Lagrangian Multipliers.

Your method is also good.

Nice use of complex nos!

max x+2y = sqrrt(5) at x = 1/[sqrrt(5)], y = 2/[sqrrt(5)].
i saw a similar sum in the video but the constraint was linear i.e. opposite of this problem.
Graphically, if we make the line x+2y=c tangent to the circle x^2 + y^2 = 1, we get the same result.