Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 6M

Year: Dec 2014

Let f = ${\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{1}}$ + ${\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{2}}$ + ${\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{3}}$ - $\boldsymbol{6}{\boldsymbol{x}}_{\boldsymbol{1}}$- 10${\boldsymbol{x}}_{\boldsymbol{2}}$ - $\boldsymbol{14}{\boldsymbol{x}}_{\boldsymbol{3}}\boldsymbol{+}\boldsymbol{103}$

Conditions to find stationary points are

$\frac{\partial f}{\partial x_1}$ = 0 $\mathrm{\to}$ $\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{-}\boldsymbol{6}\boldsymbol{=}\boldsymbol{0}$$\mathrm{\therefore }$ ${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{=}\boldsymbol{3}$

$\frac{\partial f}{\partial x_2}$ = 0 $\mathrm{\to}$ $\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{10}\boldsymbol{=}\boldsymbol{0}$$\mathrm{\therefore }$ ${\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{5}$

$\frac{\partial f}{\partial x_3}$ = 0 $\mathrm{\to}$ $\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{3}}\boldsymbol{-}\boldsymbol{14}\boldsymbol{=}\boldsymbol{0}$$\mathrm{\therefore }$ ${\boldsymbol{x}}_{\boldsymbol{3}}\boldsymbol{=}\boldsymbol{7}$

Stationary Point ${\boldsymbol{X}}_{\boldsymbol{0}}$ = (3,5,7)

Now, $\boldsymbol{\ \ \ }\frac{{\partial }^2f}{\partial {\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{1}}\boldsymbol{\mathrm{\ }}}=2$; $\frac{{\partial }^2f}{\partial x_{1\ }\partial x_2\boldsymbol{\mathrm{\ }}}=0;$ $\frac{{\partial }^2f}{\partial x_{2\ }\partial x_3\boldsymbol{\mathrm{\ }}}=0;$ $\frac{{\partial }^2f}{\partial {\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{2}}\boldsymbol{\mathrm{\ }}}=2$ $\frac{{\partial }^2f}{\partial x_{1\ }\partial x_3\boldsymbol{\mathrm{\ }}}=0;$ ; $\frac{{\partial }^2f}{\partial {\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{3}}\boldsymbol{\mathrm{\ }}}=2$

Heissian Matrix =$\left| \begin{array}{ccc} \frac{{\partial }^2f}{\partial {\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{1}}\boldsymbol{\mathrm{\ }}} & \frac{{\partial }^2f}{\partial x_{1\ }\partial x_2\boldsymbol{\mathrm{\ }}} & \frac{{\partial }^2f}{\partial x_{1\ }\partial x_3\boldsymbol{\mathrm{\ }}} \\ \frac{{\partial }^2f}{\partial x_{1\ }\partial x_2\boldsymbol{\mathrm{\ }}} & \frac{{\partial }^2f}{\partial {\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{2}}\boldsymbol{\mathrm{\ }}} & \frac{{\partial }^2f}{\partial x_{2\ }\partial x_3\boldsymbol{\mathrm{\ }}} \\ \frac{{\partial }^2f}{\partial x_{1\ }\partial x_3\boldsymbol{\mathrm{\ }}} & \frac{{\partial }^2f}{\partial x_{2\ }\partial x_3\boldsymbol{\mathrm{\ }}} & \frac{{\partial }^2f}{\partial {\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{3}}\boldsymbol{\mathrm{\ }}} \end{array} \right|$

$ = \left| \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array} \right|$

The value of determinants of principal minor are $\left|2\right|$ = 2; $\left| \begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array} \right|$ = 4; $\left| \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array} \right|=8$

Since all the determinants are positive z is minimum at ${\boldsymbol{X}}_{\boldsymbol{0}}$ = (3,5,7)

$\mathrm{\therefore } {\boldsymbol{z}}_{\boldsymbol{min}}\boldsymbol{=\ }\ 3^2+\ 5^2+\ 7^2-6\left(3\right)-10\left(5\right)-14(7) +103$

$ = 9+\ 25+49-18-50-98+103$

$= 20$