Posted by Soroban on August 25, 2002 at 02:54:25:
In Reply to: minimising posted by KK on August 25, 2002 at 02:22:39:
: How can we show that this function cannot be minimised?
: y= 2000 + 20x + 10(x)^1/2 [for x > or equal to 0]
: I tried deriving it but i always find I could minimise it because my result with the 2nd derivative was greater than 0.
The function has NO critical values, KK.
Either your derivative is incorrect,
or your subsequent algebra.
y' = 20 + 5x^(-1/2) = 20 + 5/sqrt(x)
The sqrt is always taken as positive.
20 plus a positve fraction cannot equal 0.
Post a Followup