# Multivariable optimization question. Find three positive real numbers whose sum is one and the sum of their squares is a minimum.

Multivariable optimization question. Find three positive real numbers whose sum is one and the sum of their squares is a minimum.
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Let the three positive real numbers be x,y and z.
then, $x+y+z=1$
let sum of squares be:
$f\left(x,y,z\right)={x}^{2}+{y}^{2}+{z}^{2}$
put z in above function
$g\left(x,y\right)={x}^{2}+{y}^{2}+{\left(1-\left(x+y\right)\right)}^{2}$
$={x}^{+}{y}^{2}+1+{\left(x+y\right)}^{2}-2\cdot 1\cdot \left(x+y\right)$
$={x}^{2}+{y}^{2}+1+{2}^{2}+{y}^{2}+2xy-2x-2y$
$=2{x}^{2}+2{y}^{2}+2xy-2x-2y+1$
Partially differenting the function g(x,y) and finding the critical points:
$\frac{\partial g}{\partial x}=4x+2y-2=0$
$\frac{\partial g}{\partial y}=4y+2x-2=0$
From the above two equtioan, we get:
$x=\frac{1}{3},y=\frac{1}{3}$
Putting the value of x and y in $x+y+z=1$, we get, $z=\frac{1}{2}$
Using the second derivative test:
${g}_{×}=4,$
${g}_{yy}=4,$
${g}_{xy}4x+2=4\left(\frac{1}{3}\right)+2=\frac{10}{3}$
$D={g}_{xx}\cdot {g}_{yy}-{g}_{xy}^{2}$
$=4\cdot 4-{\left(\frac{10}{3}\right)}^{2}$
$=16-\frac{100}{9}$
$=\frac{44}{9}>0$
Since ${g}_{xx}>0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}D>0,p\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$ is a local minimal
Therefore, the sum of squarers will be minimom if $x=y=z=\frac{1}{3}$ Hence, $x=\frac{1}{3},y=\frac{1}{3},z=\frac{1}{3}.$