Question

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

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

2021-02-20

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