Re: finding minimum
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Tue, 22 Nov 2005 17:21:27 +0000 (UTC)
In article <1132679627.949875.180180@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mathforsweta@xxxxxxxxx <mathforsweta@xxxxxxxxx> wrote:
>Hello Everyone,
>
>
>I have m=min(a,1/2a)..I want to find the largest of m.
Over what interval? Presumably, a>0?
>I differentialted 1/2a with respect to a and set it equal to zero. That
>is what we do when we want to find maximum or minimum.
>
>the true answer is a=1/sqrt(2)
>
>May be I am doing something wrong?
>
>Can somebody please help me with this problem?
First, you need to figure when m = a, and when m=1/(2a) (I assume that
is what you mean by 1/(2a) ). That is, you
need to know when a-(1/2a) is positive, and when it is
negative. Rewriting this, we have
a - (1/(2a)) = (2a^2 - 1)/2a.
This is positive if both 2a and 2a^2-1 are positive, or if both are
negative. It is negative if 2a^2-1 is positive and 2a negative, or if
2a^2-1 is negative and 2a positive.
So: 2a is positive if a>0, and negative if a<0.
2a^2 - 1 is positive if a^2 > 1/2, which happens when a>1/sqrt(2), or
when a< - 1/sqrt(2). So it is positive if a>1/sqrt(2) or if
a<-1/sqrt(2); and it is negative if a is between -1/sqrt(2) and
1/sqrt(2).
So:
a - (1/(2a)) is positive if a > 1/sqrt(2), or if a is between
-1/sqrt(2) and 0. And it is negative if a is between 0 and 1/sqrt(2),
or if a is less than -1/sqrt(2).
So another way to write m is:
/ a if a<= -1/sqrt(2) or 0<a<= 1/sqrt(2)
m = {
\ 1/(2a) if -1/sqrt(2) <= a < 0 or a>1/sqrt(2).
You want to find the maximum of m. Since a is increasing on
(-infty,infty), and 1/(2a) is increasing on (-infty,0) and decreasing
on (0,infty), it should be s imple matter to check the maximum value
of m in the different intervals in question; and then the largest
maxima will be the global maximum.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx
.
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