Can you calculate least value?

We can easily notice that points we are asked about lie on the parabola $y=x^2$ , for $x=\{1,2,3,4,5\} \rightarrow y=\{1,4,9,16,25\}$ .

Let the coordinates of P be: $P=(x_p,y_p)$ .

So now we can write $k$ as a function dependent on $x_p$ and $y_p$ , namely:

$k(x_p,y_p)= \sum _{ i=1 }^{ 5 }{ \left[ { \left( { x }_{ p }-{ x }_{ i } \right) }^{ 2 }+{ \left( { y }_{ p }-{ y }_{ i } \right) }^{ 2 } \right] } = \sum _{ i=1 }^{ 5 }{ \left[ { \left( { x }_{ p }-i \right) }^{ 2 }+{ \left( { y }_{ p }-{ i }^{ 2 } \right) }^{ 2 } \right] }$

Expanding, we get:

$k(x_p,y_p)= { ( { x }_{ p }-1) }^{ 2 }+ { ( { y }_{ p }-1) }^{ 2 } + { ( { x }_{ p }-2) }^{ 2 }+ { ( { y }_{ p }-4) }^{ 2 } +$ ${ ( { x }_{ p }-3) }^{ 2 }+ { ( { y }_{ p }-9) }^{ 2 } + { ( { x }_{ p }-4) }^{ 2 }+ { ( { y }_{ p }-16) }^{ 2 } + { ( { x }_{ p }-5) }^{ 2 }+ { ( { y }_{ p }-25) }^{ 2 } =$ $=5{{x}_{p}}^{2}+5{{y}_{p}}^{2}-30{x}_{p}-110{y}_{p}+1034$

Now we need to find critical points of the function $k(x_p,y_p)$ :

$\frac{\partial k(x_p,y_p)}{\partial x_p}= 10x_p-30=0 \Rightarrow x_p=3$

$\frac{\partial k(x_p,y_p)}{\partial y_p}= 10y_p-110=0 \Rightarrow y_p=11$

Substituting, we get:

$k(3,11)= 384$

$Note:$ generally, we should also make sure that the found critical point is the minimum. To do so we need to calculate determinant of a matrix in our critical point:

$\begin{bmatrix} \frac { { \partial }^{ 2 }k({ x }_{ p },{ y }_{ p }) }{ { \partial { x }_{ p } }^{ 2 } } & \frac { { \partial }^{ 2 }k({ x }_{ p },{ y }_{ p }) }{ { \partial { x }_{ p } }\partial { y }_{ p } } \\ \frac { { \partial }^{ 2 }k({ x }_{ p },{ y }_{ p }) }{ { \partial { y }_{ p }\partial { x }_{ p } } } & \frac { { \partial }^{ 2 }k({ x }_{ p },{ y }_{ p }) }{ { \partial { y }_{ p } }^{ 2 } } \end{bmatrix}$

Which in our problem takes the form:

$= \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$

So the determinant is:

$H(3,11)=\begin{vmatrix} 10 & 0 \\ 0 & 10 \end{vmatrix}=100 >0$

Because this determinant is greater than 0, point $(3,11)$ is a minimum.