<s>[INST] Find the point on the graph of $z= x^{2} +y^{2} +10$ nearest to the plane $x+2y-z=0$. Find the point on the graph of $z= x^{2} +y^{2} +10$ nearest to the plane $x+2y-z=0$.
So, any point on the given surface will be $(x,y,x^{2} +y^{2} +10)$ .
I need to minimize the function $(x +2y-x^{2}-y^{2}-10)/(\sqrt{6})$ 
The only critical point is $(1/2,1)$.
But this point gives maximum of the function.
How would I find the nearest point.
What I think is that, I should change the sign of the function, to keep the distance positive, then I'll get the same critical point, but the value will be minimum. So I'll get the nearest point. 
In this case if I am asked to find the maximum distance, what it should be then$?$
 [/INST] Note that you have to minimize the function $$ \frac{ |x +2y-x^{2}-y^{2}-10|}{\sqrt{6}} $$In your post, you does not take absolute value.
Observe that $$x^2 + y^2 -x - 2y + 10=(x-\frac{1}{2})^2 + (y-1)^2 + \frac{35}{4} $$ which is always positive. so $$ \frac{ |x +2y-x^{2}-y^{2}-10|}{\sqrt{6}} = \frac{ x^2+y^2-x-2y+10}{\sqrt{6}} \geq \frac{35}{4\sqrt6}
 $$ It attains its minimum at $(x, y)=(\frac 12, 1)$.
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