Re: minimising


Posted by Soroban on August 27, 2002 at 02:18:04:

In Reply to: Re: minimising posted by kk on August 26, 2002 at 20:22:05:

:But how does it prove that the equation
: y=(2000+20x+10x^1/2)/x cannot be minimised (i.e. the function does not have a convex). I thought we have to take the second derivative because for a convex to occur f"(x)>0.

Please listen carefully, KK.

The function had NO critical values.
(We set the first derivative equal to 0
and could not solve for x.)
Therefore, there are NO maximums or minimums.

Why do we take the second derivative?
To test the Critical Values for max or min.
But there are NO critical values to test.

[Your flashlight doesn't work. I tell you that
there are no batteries in it. And you insist on
trying a battery-tester?]

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