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

Marks: 8M

Year: May 2015

Let $$f(x_1,x_2) = x^2_1 +x^2_2;h_1= x_1+x_2-4;\ h_2= 2x_1+x_2-5$$

Let $\lambda$ be Lagrangian multiplier

Lagrangian function L = f - $h_1{\mathrm{\lambda}}_1--h_2{\mathrm{\lambda}}_2$

L = $(\mathrm{\ }x^2_1$ $+$ $x^2_2)$ $-$ ${\mathrm{\lambda}}_1$ ($x_1+x_2-4)\ --$ ${\mathrm{\lambda}}_2$( $2x_1+x_2-5)$

L = $\mathrm{\ \ }x^2_1$ $+$ $x^2_2$ $--$ ${\mathrm{\lambda}}_2$ $x_1-{\mathrm{\lambda}}_2x_2+{\mathrm{\lambda }}_24\ --$ ${\mathrm{\lambda}}_22x_1-{\mathrm{\lambda}}_2x_2+{\mathrm{\lambda}}_25$

Kuhn-Tucker conditions are:

$$1)\frac{\partial L}{\partial x_1} = 0 \mathrm{\to}\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{-}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}\boldsymbol{-}\boldsymbol{2}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{0}$$

$$2)\frac{\partial L}{\partial x_2} = 0 \mathrm{\to}\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}\boldsymbol{-}{\boldsymbol{\mathrm{\lambda }}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{0}$$

$$\boldsymbol{3}\boldsymbol{)\ }{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}\boldsymbol{h} ({\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{)}\boldsymbol{=}\boldsymbol{0}\mathrm{\to}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}({\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{4}\boldsymbol{)\ }\boldsymbol{=}\boldsymbol{0}$$ $$\boldsymbol{4}\boldsymbol{)\ }{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}\boldsymbol{h} ({\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{)} \boldsymbol{=}\boldsymbol{0}\boldsymbol{\mathrm{\ \ \ }}\mathrm{\to }\boldsymbol{\mathrm{\ \ \ }}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}\boldsymbol{\mathrm{\ (}}\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{5}\boldsymbol{)\ }\boldsymbol{\mathrm{\ \ \ \ }}\boldsymbol{=}\boldsymbol{0}\boldsymbol{\mathrm{\ \ \ \ }}$$ $$\boldsymbol{5}\boldsymbol{)\ }{\boldsymbol{h}}_{\boldsymbol{1}}({\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{)}\mathrm{\le }\mathrm{\ } \boldsymbol{0} \mathrm{\to }\mathrm{\ \ \ \ \ \ }{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-} \boldsymbol{4} \mathrm{\le }\boldsymbol{0}$$ $$\boldsymbol{6}\boldsymbol{)\ }{\boldsymbol{h}}_{\boldsymbol{2}}({\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{)}\mathrm{\le }\mathrm{\ } \boldsymbol{0} \mathrm{\to }\mathrm{\ \ \ \ \ \ }\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{5} \mathrm{\le }\boldsymbol{0}$$ $$\boldsymbol{7}\boldsymbol{)\ }{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}},{\boldsymbol{\mathrm{\lambda }}}_{\boldsymbol{2}},{\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}\mathrm{\ge} 0$$

From Kuhn-Tucker conditions (2) (4) following 4 cases arises

Case 1. ${\mathrm{\lambda}}_1={\mathrm{\lambda}}_2=0$

From (1), $2x_1-0-0$ = 0 $$\mathrm{\ }\mathrm{\therefore }\mathrm{\ }x_1=0$$ From (2), $2x_2-0-0$= 0 $$\mathrm{\therefore }\mathrm{\ }x_2=0$$ Substitute $x_1\ and\ x_2\ in\ (5)$

LHS =$\ 0+0-4=-4\lt0$

Substitute $x_1\ and\ x_2\ in \ (6)$

LHS =$\ 0+0-5=-5\lt0$

Values${\mathrm{\lambda}}_1,{\mathrm{\lambda}}_2,x_1,{\&\ x}_2$ satisfies Kuhn-Tucker conditions

Stationary Point $X_0$ = $(0,0)$

Z is maximum $$\mathrm{\therefore } {\boldsymbol{z}}_{\boldsymbol{max}}\boldsymbol{=\ }\ \ 0^2+0^2$$ $$ = 0$$ Case 2. ${\mathrm{\lambda}}_1\neq 0,\ \ {\mathrm{\lambda}}_2=0$

From (3), $x_1+x_2-4$= 0

$$\mathrm{\ }\mathrm{\therefore }\mathrm{\ }x_1+x_2=4 \mathrm{\to }.......(8)$$

From (1), $2x_1$ $-{\mathrm{\lambda}}_1-0$= 0

$$\mathrm{\therefore }\mathrm{\ }2x_1+{\mathrm{\lambda}}_1= 0 \mathrm{\to }\mathrm{\ }...(9)$$

From (2), $2x_2-{\mathrm{\lambda}}_1-0$= 0

$$\mathrm{\therefore }\mathrm{\ }2x_2-{\mathrm{\lambda}}_1= 0 \mathrm{\to }\mathrm{\ }....(10)$$

Solving (8) (9) and (10) simultaneously, we get $${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{=}\boldsymbol{2}\boldsymbol{;} {\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{2}; {\boldsymbol{\mathrm{\lambda }}}_{\boldsymbol{1}}\boldsymbol{=}\boldsymbol{4}$$ substitute $x_1,x_2$ in (6)

LHS = $2\left(2\right)+3-5=1$ $\nleq 0$

Values $x_1\&\ x_2$ does not satisfies Kuhn-Tucker conditions (6) $$\mathrm{\therefore }\mathrm{We\ Reject\ this\ case}$$ Case 3. ${\mathrm{\lambda}}_1=0,\ \ {\mathrm{\lambda}}_2\neq 0$

From (4), $2x_1+x_2-5\ \mathrm{\ \ \ \ }=0\mathrm{\ \ \ \ }$

$$\mathrm{\ }\mathrm{\therefore }\mathrm{2}x_1+x_2=5\ \mathrm{\ \ \ \ } \mathrm{\to }........(11)$$

From (1) $2x_1$ $-{\mathrm{2}\mathrm{\lambda}}_2-0$= 0

$$\mathrm{\therefore }\mathrm{\ }2x_1-2{\mathrm{\lambda}}_2= 0 \mathrm{\to }\mathrm{\ }.......(12)$$

From (2), $2x_2-{\mathrm{\lambda}}_2-0$= 0

$$\mathrm{\therefore }2x_2-{\mathrm{\lambda}}_2= 0 \mathrm{\to }\mathrm{\ }......(13)$$

Solving (11)(12) and (13) simultaneously, we get $$x_1=2; x_2=1; {\mathrm{\lambda}}_2=2$$ substitute $x_1,x_2$ in (5)

LHS = $2+1-4=-1$ $\lt0$

Values${\mathrm{\lambda}}_1,{\mathrm{\lambda}}_2,x_1,{\&\ x}_2$ satisfies Kuhn-Tucker conditions

Stationary Point $X_0$ = $(2,1)$

Z is maximum $$\mathrm{\therefore } z_{max}=\ \ \ 2^2+1^2 = 5$$

Case 4. $${\mathrm{\lambda}}_1\neq 0,\ \ {\mathrm{\lambda}}_2\neq 0$$

From (3), $x_1+x_2-4$= 0

$$\mathrm{\ }\mathrm{\therefore }\mathrm{\ }x_1+x_2=4 \mathrm{\to }.......(14)$$

From (4), $$2x_1+x_2-5\mathrm{\ \ \ \ }=0\mathrm{\ \ \ \ }$$

$$\mathrm{\ }\mathrm{\therefore }2x_1+x_2=5\ \mathrm{\ \ \ \ } \mathrm{\to }....................(15)$$

Solving (14)and (15) simultaneously, we get $$x_1=1; x_2=3;$$ From (1), $2x_1-{\mathrm{\lambda}}_1$ $-{\mathrm{2}\mathrm{\lambda}}_2-0$= 0 $$\mathrm{\therefore }\mathrm{\ }2(1)-{\mathrm{\lambda}}_1-2{\mathrm{\lambda}}_2= 0$$

$${\mathrm{\lambda}}_1+2{\mathrm{\lambda}}_2=2 \mathrm{\to }\mathrm{\ }.....(16)$$

From (2), $2x_2-{\mathrm{\lambda}}_1\mathrm{\ }-{\mathrm{\lambda}}_2-0$= 0 $$\mathrm{\therefore }2(3)-{\mathrm{\lambda}}_1-{\mathrm{\lambda}}_2= 0 $$

$${\mathrm{\lambda}}_1+2{\mathrm{\lambda}}_2\mathrm{=6\ }\mathrm{\to }\mathrm{\ } ......(17)$$

Solving (16 & (17) we get ${\mathrm{\lambda}}_1$ $\mathrm{=10\ }$ ${\mathrm{\lambda}}_2\mathrm{=-4}$

Value of ${\mathrm{\lambda}}_2$ do not satisfy Kuhn-Tucker conditions $$\mathrm{\therefore }\mathrm{We\ Reject\ this\ case}$$

Hence from Case 1,2,3 & 4 We conclude that

$\mathrm{\therefore }$ $z_{max}$ = $5$ at $x_1=2;$ $x_2=1$; ${\mathrm{\lambda}}_1=2$