# the nearest point sort of problems

• March 4th 2009, 05:10 PM
the nearest point sort of problems
It is a minimization problem and i don't really understand them. Please give me some reasonable solution on the following one:

Find the point (
x, y) belong toR2 on the graph of y = x^(1\2) nearest to the point (4, 0).

Thanks X 100000
• March 4th 2009, 08:57 PM
Soroban

Quote:

Find the point $(x,y)$ on the graph of $y\,=\,x^{\frac{1}{2}}$ nearest to the point $(4, 0).$
Perhaps a sketch will help . . .
Code:

        |         |            *         | (x,y) *         |  o         | *  \         |*    \d         |      \     - - * - - - o- - - - -         |    (4,0)

The point has coordinates $(x,y) \,=\,\left(x,\sqrt{x}\right)$

We want its distance from $(4,0)$ to be a minimum.

The distance is: . $d \;=\;\sqrt{(x-4)^2 + \left(\sqrt{x} - 0\right)^2}$

. . So we have: . $d \;=\;\left(x^2-7x+16\right)^{\frac{1}{2}}$

Then: . $d\,' \;=\;\tfrac{1}{2}\left(x^2-7x+16\right)^{-\frac{1}{2}}(2x-7) \quad\Rightarrow\quad \frac{2x-7}{2\sqrt{x^2-7x+16}} \;=\;0$

Hence: . $2x-7\:=\:0 \quad\Rightarrow\quad x \:=\:\frac{7}{2}$

And: . $y \:=\:\sqrt{\frac{7}{2}} \:=\:\frac{\sqrt{14}}{2}$

Therefore, the nearest point is: . $\left(\frac{7}{2},\:\frac{\sqrt{14}}{2}\right)$