I need to find the closest point of the graph $y=\sqrt x$ to the point $(4,0)$.
This is how I understand it:
I assume that triangle with the sides $a$, $x$ and $x/2$ and the largest triangle are similar triangles. From that I conclude that the smaller triangle's one side is $x/2$.
The closest point on the graph from $(4,0)$ must be in the intersection of the lines $a$, $b$ and $c$. $$a = \sqrt{x^2+(x/2)^2},\quad b = \sqrt{x^2 + (2-x/2)^2},\quad c = \sqrt{(4-x)^2+(x/2)^2}$$ by the Pythagorean theorem.
I have to minimize the sum $a+b+c$ by diferentiating and then solving for $x$, and then subtracting $x$ from 4 would get the abscissa of the wanted point.
Solving with wolframalpha $x \approx 0.57$, so the abscissa is $4-0.57=3.43$, ordinate is $0.57/2=0.285$.
Is this correct?
Best Answer
The distance from a point on the graph $y=\sqrt{x}$ to the point $(4,0)$ is $$\sqrt{(x-4)^2+(\sqrt{x}-0)^2}.$$ This will be minimized precisely when $$(x-4)^2+(\sqrt{x}-0)^2=x^2-8x+16+x=x^2-7x+16$$ is minimized, since $\sqrt{\cdot}$ is an increasing function. Do you know how to find $x\geq 0$ that minimizes $x^2-7x+16$?