I would like to find the points at which $x$ is maximum for any $y$ and $z$, $y$ is maximum for any $x$ and $z$, and $z$ is maximum for any $x$ and $y$.
I have the following 3 variable ($x$, $y$ and $z$) system of equations; $$ c = \frac{(x-m)^2}{m} + \frac{(y-k)^2}{k} + \frac{(z-j)^2}{j}$$ and; $$ x+y+z=m+k+j=N$$
where $c$, $m$, $k$, $j$ and $N$ are all postive constants. Furthermore, $x$, $y$ and $z$ are constrained to positive values. I have an inkling that partial derivatives can produce an answer but i can't get there. Keep in mind that 'novice' would be a generous description of my expertise.
Edit : Working solution as per @Avi 's comment i am attempting to use Lagrange multiplier to solve this problem; $$ F(x,y,z) = \frac{(x-m)^2}{m} + \frac{(y-k)^2}{k} + \frac{(z-j)^2}{j} -c$$
$$ F(x,y,z) = \frac{(x^2-2mx+m^2)}{m} + \frac{(y^2-2ky+k^2)}{k} + \frac{(z^2-2jz+j^2)}{j} -c$$ $$ F(x,y,z) = \frac{x^2}{m} + \frac{y^2}{k} + \frac{z^2}{j} - 2x-2y-2z -c + m+k+j$$
So then considering my Lagrange multiplier to be; $$F(\lambda) = x+y+z-N$$
I have; $$ F(x,y,z,\lambda) = \frac{x^2}{m} + \frac{y^2}{k} + \frac{z^2}{j} - 2x-2y-2z -c + m+k+j + \lambda(x+y+z-N)$$
Setting partial derivatives to $0$, i get; $$F_x=\frac{2x}{m}-2+\lambda=0$$ $$F_y=\frac{2y}{k}-2+\lambda=0$$ $$F_x=\frac{2z}{j}-2+\lambda=0$$ $$F_\lambda = x+y+z-N=0$$
Solving the first 3 equations for $x$,$y$ and $z$, i get; $$x=m-\frac{m\lambda}{2}$$ $$y=k-\frac{k\lambda}{2}$$ $$z=j-\frac{j\lambda}{2}$$ and plugging these into $F_\lambda$ above;
$$F_\lambda=m+k+j - N- \frac{\lambda(m+k+j)}{2}=0$$ solving for $\lambda$ i get; $$\lambda=\frac{2(m+k+j-N)}{m+k+j}$$ and since $m+k+j=N$; then $\lambda=0$ and; $$x=m, y=k, z=j$$
So the function $F(x,y,z)$ is at a minimum when $x=m, y=k, z=j$, which makes perfect sense. But what about maxima?