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Задача на мінімум у лінійному програмуванні має вигляд
f({{x}_{1}},{{x}_{2}},...,{{x}_{n}})={{c}_{1}}{{x}_{1}}+{{c}_{2}}{{x}_{2}}+...+{{c}_{n}}{{x}_{n}}\to \min
\left\{ \begin{matrix} a_{11}{x}_{1}+{a}_{12}{x}_{2}+...+{{a}_{1n}}{{x}_{n}}\ge {{b}_{1}}, \\ {a}_{21}{{x}_{1}}+{{a}_{22}}{{x}_{2}}+...+{{a}_{2n}}{{x}_{n}}\ge {{b}_{2}}, \\ ... \\ {{a}_{m1}}{{x}_{1}}+{{a}_{m2}}{{x}_{2}}+...+{{a}_{mn}}{{x}_{n}}\ge {{b}_{m}}, \\\end{matrix} \right.
\left\{ \begin{matrix}
a_{11}{x}_{1}+{a}_{12}{x}_{2}+...+{{a}_{1n}}{{x}_{n}}\ge {{b}_{1}}, \\
{a}_{21}{{x}_{1}}+{{a}_{22}}{{x}_{2}}+...+{{a}_{2n}}{{x}_{n}}\ge {{b}_{2}}, \\
... \\
{{a}_{m1}}{{x}_{1}}+{{a}_{m2}}{{x}_{2}}+...+{{a}_{mn}}{{x}_{n}}\ge {{b}_{m}}, \\
\end{matrix} \right.
f({{x}_{1}},{{x}_{2}},...,{{x}_{n}})={{c}_{1}}{{x}_{1}^2}+{{c}_{2}}{{x}_{2}^2}+...+{{c}_{n}}{{x}_{n}^2}\to \min
\left\{ \begin{matrix} a_{11}{x}_{1}+{a}_{12}{x}_{2}+...+{{a}_{1n}}{{x}_{n}}\ge {{b}_{1}}, \\ {a}_{21}{{x}_{1}}+{{a}_{22}}{{x}_{2}}+...+{{a}_{2n}}{{x}_{n}}\ge {{b}_{2}}, \\ ... \\ {{a}_{m1}}{{x}_{1}}+{{a}_{m2}}{{x}_{2}}+...+{{a}_{mn}}{{x}_{n}}\ge {{b}_{m}}, \\ \end{matrix} \right.
\left\{ \begin{matrix} a_{11}{x}_{1}+{a}_{12}{x}_{2}+...+{{a}_{1n}}{{x}_{n}}\ge {{b}_{1}}, \\ {a}_{21}{{x}_{1}}+{{a}_{22}}{{x}_{2}}+...+{{a}_{2n}}{{x}_{n}}\ge {{b}_{2}}, \\ ... \\ {{a}_{m1}}{{x}_{1}^2}+{{a}_{m2}}{{x}_{2}^2}+...+{{a}_{mn}}{{x}_{n}^2}\ge {{b}_{m}}, \\ \end{matrix} \right.
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