1. ## optimization

find the point on the curve y=x^1/2 nearest to the point (c,0) if c(> and =)1/2

2. Originally Posted by Joyce
find the point on the curve y=x^1/2 nearest to the point (c,0) if c(> and =)1/2
We need to minimize the distance between the points $\displaystyle (x, x^{1/2})$ and (c, 0).

The distance between two the two points is:
$\displaystyle d = \sqrt{(x - c)^2 + (x^{1/2} - 0)^2}$

Note that if we square the distance:
$\displaystyle d^2 = (x - c)^2 + (x^{1/2} - 0)^2$
we still find the minimum x in the same place with a much easier computation.

So now take your x derivative with respect to the $\displaystyle d^2$ equation, set it equal to 0, etc.

Not sure why you've got the $\displaystyle c \geq 1/2$ part, though.

Edit: Ahhhh, $\displaystyle c \geq 1/2$ because the domain of the function is only positive x.

-Dan