Find the minimum distance from the point

Question

Find the minimum distance from the point $ \ (5,0) \ $ to the curve $ \ y=\sqrt x+2 \ $.

Answer:

Let $ \ (x,y) \ $ be the closest point on $ \ y=\sqrt x+2 \ $ from $ (5,0) $ .

Then the distance is given by

$ d(x,y)=\sqrt{(x-5)^2+y^2} \ $

We will minimize the function $ \ g(x,y)=(x-5)^2+y^2 \ $

replacing $ y \ \ by \ \ \sqrt x+2 \ $ , we get

$ g(x)=(x-5)^2+(\sqrt x+2)^2 \ $

The extreme points \ are

$ f'(x)=0 \\ 2(x-5)+\frac{\sqrt x+2}{\sqrt x} =0 \\ 2x \sqrt x-9 \sqrt x+2=0 $

This becomes complicated . I am unable to calculate the closest point.

Help me out

Answer 1

Now, let $\sqrt{x}=t$.

Thus,$$2x\sqrt{x}-9\sqrt{x}+2=2t^3-9t+2=2t^3-4t^2+4t^2-8t-t+2=$$ $$=(t-2)(2t^2+4t-1).$$ Thus, $x_{min}=4$ and $x_{max}=\left(\sqrt{1.5}-1\right)^2.$

Also, we need to check, what happens for $x=0$.

Answer 2

write your last equation as $$2x-10+1+\frac{2}{\sqrt{x}}=0$$ and Isolate the term with the square root: $$\frac{2}{\sqrt{x}}=-2x+9$$ squaring gives $$\frac{4}{x}=(-2x+9)^2$$ can you finish? factorizing the last equation gives $$- \left( x-4 \right) \left( 4\,{x}^{2}-20\,x+1 \right) =0$$