# Show that a point is closest to another point with extreme value theorem

Show there is a point of the plane $$\{x \in \mathbb{R^3} \mid x_1 + 2x_2 + 3x_3 = 13\}$$ closest to the point $$(1, 1, 1)$$.

Let a function $$f: A \rightarrow \mathbb{R}$$ defined for all $$x \in A$$ by $$f(x_1, x_2, x_3) = x_1 + 2x_2 + 3x_3 - 13$$.

We can take the domain $$A = x_1 + 2x_2 + 3x_3 = 13 \cap B[(1, 1, 1), r]$$.

If we take $$r = (1, 1, \frac{10}{3})$$, the plane intersects the ball.

So, this domain is bounded and closed. We show that $$f$$ is continuous.

So, by the extreme value theorem, there is $$a \in A$$ and $$b \in A$$ such that for all $$x \in A$$ : $$f(a) \le f(x) \le f(b)$$.

We showed that there is a point $$a \in A$$ that is closest to the point $$(1, 1, 1)$$.

The basic idea behind this proof is that a plane is closed but not bounded. So we intersect the set with a closed ball then we can use the extreme value theorem.

Is it correct ?

• Hm.The ball's radius $r$ should be a nonnegative number and not a point (you wrote $r=(1,1,10/3)$ ) – Maksim Mar 18 at 14:59