# Re: minimising

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Posted by T.Gracken on August 25, 2002 at 09:12:25:

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.

I'm not quite sure what you mean by "minimising the function", but if you want to determine if the funcion has a minimum...

There is no need to use the second derivative.

determine y': y'=20 + 5x-1/2

now, critcal values for y (that is, domain values of y for which y' is either 0 or undefined) are at x=0.

Note that y' is positive for all x in the domain of y' (0,inf), so the function is strictly increasing on (0,inf).

This leads to the function having a minimum when x=0.

so 2000 (the value of y when x=0) is an (absolute) minimum.

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