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

Marks: 6M

Year: Dec 2015

Let $$f(x_1,x_2) = 6x^2_1+5x^2_2\ ; h = x_1+5x_2-7$$ and $\lambda$ be Lagrangian multiplier

Lagrangian function L = f - $\lambda$h

L = $(6$ $x^2_1$ $+$ $5x^2_2)$ $-$ $\lambda$($x_1+5x_2-7)$

L = $6$ $x^2_1$ $+$ $5x^2_2$ $-$ $\lambda$$x_1-\mathrm{\ }\mathrm{\lambda }5x_2+\mathrm{\ }\mathrm{\lambda }7)$ Conditions to find stationary points are $\frac{\partial L}{\partial x_1}$ = 0 $\mathrm{\to}$ $12x_1-\mathrm{\lambda }=0$ $\mathrm{\therefore }$ $12x_1-\mathrm{\lambda }=0$ $\mathrm{\to}$ (1)

$\frac{\partial L}{\partial x_2}$ = 0 $\mathrm{\to}$ $10x_2-5\mathrm{\lambda }=0$ $\mathrm{\therefore }$ $10x_2-5\mathrm{\lambda }=0$ $\mathrm{\to}$(2)

$\frac{\partial L}{\partial \mathrm{\lambda }}$ = 0 $\mathrm{\to}$ $-x_1-5x_2+7=0$ $\mathrm{\therefore }$ $x_1+{5x}_2=7$ $\mathrm{\to}$(3)

Solving (1) (2) and (3) simultaneously $$x_1=\frac{7}{31}\ ; x_2=\frac{42}{31}; \mathrm{\lambda }\mathrm{=}\frac{84}{31} ;$$

Stationary Point $X_0$ = ($\frac{7}{31}$,$\ \frac{42}{31}$)

Now, $\ \ \ \frac{{\partial }^2L}{\partial x^2_1\mathrm{\ }}=12$; $\frac{{\partial }^2L}{\partial x_{1\ }\partial x_2\mathrm{\ }}=0;$ $\frac{{\partial }^2L}{\partial x^2_2\mathrm{\ }}=10$ ; $\frac{\partial h}{\partial x_1}=1;$ $\frac{\partial h}{\partial x_2}=5$

$\boldsymbol{\Delta}$ = $\left| \begin{array}{ccc} 0 & \frac{\partial h}{\partial x_1} & \frac{\partial h}{\partial x_2} \\ \frac{\partial h}{\partial x_1} & \frac{{\partial }^2L}{\partial {\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{1}}\boldsymbol{\mathrm{\ }}} & \frac{{\partial }^2L}{\partial x_{1\ }\partial x_2\boldsymbol{\mathrm{\ }}} \\ \frac{\partial h}{\partial x_2} & \frac{{\partial }^2L}{\partial x_{1\ }\partial x_2\boldsymbol{\mathrm{\ }}} & \frac{{\partial }^2L}{\partial {\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{2}}\boldsymbol{\mathrm{\ }}} \end{array} \right|$ $$ = \left| \begin{array}{ccc} 0 & 1 & 5 \ 1 & 12 & 0 \ 5 & 0 & 10 \end{array} \right|$$ $$= 0-1\left(10-0\right)+5\left(0-60\right)$$ $$= -310$$ $$= -310<0$$

Since $\Delta$ is negative Z is minima at $X_0$ = ($\frac{7}{31}$,$\ \frac{42}{31}$) $$\mathrm{\therefore } z_{min}=\ \ {\ {6(\frac{7}{31})}^2+5(\frac{42}{31})}^2$$ $$ = \frac{294}{31}$$