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

Marks: 8M

Year: May 2014

$ \text{Let} \ \ \ f({x}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}},{\boldsymbol{x}}_{\boldsymbol{3}}\boldsymbol{)} = \boldsymbol{-}{\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{1}}\boldsymbol{-}{\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{2}}\boldsymbol{-}{\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{3}}\boldsymbol{+}\boldsymbol{4}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{6}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{;} {\boldsymbol{h}}_{\boldsymbol{1}} = {\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{2}\boldsymbol{;}\boldsymbol{\ }{\boldsymbol{h}}_{\boldsymbol{2}} = \boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{3}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{12}$

Let $\boldsymbol{\lambda}$ be Lagrangian multiplier

Lagrangian function L = f - ${\boldsymbol{h}}_{\boldsymbol{1}}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}-{\boldsymbol{h}}_{\boldsymbol{2}}{\boldsymbol{\mathrm{\lambda }}}_{\boldsymbol{2}}$

L = $\boldsymbol{(}\boldsymbol{\mathrm{\ }}\boldsymbol{-}\boldsymbol{\mathrm{\ }}{\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{1}}\boldsymbol{\mathrm{\ }}\boldsymbol{-}\boldsymbol{\mathrm{\ }}{\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{2}}\boldsymbol{-}{\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{3}}\boldsymbol{+}\boldsymbol{4}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{6}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{)}$$-$${\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}$ (${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{2}\boldsymbol{)\ }-$${\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}$($\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{3}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{12}\boldsymbol{)}$

L = $\boldsymbol{\mathrm{\ }}\boldsymbol{-}\boldsymbol{\mathrm{\ }}{\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{1}}\boldsymbol{\mathrm{\ }}\boldsymbol{-}\boldsymbol{\mathrm{\ }}{\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{2}}\boldsymbol{-}{\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{3}}\boldsymbol{+}\boldsymbol{4}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{6}{\boldsymbol{x}}_{\boldsymbol{2}}$${\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}$${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{-}{{\boldsymbol{\mathrm{\lambda }}}_{\boldsymbol{1}}\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{+}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}\boldsymbol{2}-$${\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{-}{\boldsymbol{\mathrm{\lambda }}}_{\boldsymbol{2}}\boldsymbol{3}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{+}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}\boldsymbol{12}$ $ \\ $

Kuhn-Tucker conditions are:

1) $\frac{\partial L}{\partial x_1} = 0 \mathrm{\to} \ \ -\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{4}\boldsymbol{-}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}\boldsymbol{-}\boldsymbol{2}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{0}$

2) $\frac{\partial L}{\partial x_2} = 0 \mathrm{\to} \ -\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{+}\boldsymbol{6}\boldsymbol{-}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}\boldsymbol{-}\boldsymbol{3}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{0}$

3) $\frac{\partial L}{\partial \boldsymbol{\mathrm{\lambda}}} = 0 \mathrm{\to} \boldsymbol{-}\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{3}}\boldsymbol{=}\boldsymbol{0} \ \ \mathrm{\therefore }\ \ {\boldsymbol{x}}_{\boldsymbol{3}}\boldsymbol{=}\boldsymbol{0}$

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

5) ${\boldsymbol{h}}_{\boldsymbol{2}}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}\boldsymbol{\mathrm{(}}{\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}},{\boldsymbol{x}}_{\boldsymbol{3}}\boldsymbol{)}\boldsymbol{\mathrm{\ }}\boldsymbol{=}\boldsymbol{0}\boldsymbol{\mathrm{\ \ \ }}\mathrm{\to }\boldsymbol{\mathrm{\ \ \ }}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}\boldsymbol{\mathrm{\ (}}\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{3}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{12}\boldsymbol{)\ }\boldsymbol{\mathrm{\ \ \ \ }}\boldsymbol{=}\boldsymbol{0}\boldsymbol{\mathrm{\ \ \ \ }}$

6) ${\boldsymbol{h}}_{\boldsymbol{1}}$(${\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}},{\boldsymbol{x}}_{\boldsymbol{3}}\boldsymbol{)}\mathrm{\le}\mathrm{\ }\boldsymbol{0}$$\mathrm{\to }\mathrm{\ \ \ \ \ \ }{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{2}$$\mathrm{\le }\boldsymbol{0}$ $ \\ $

7) ${\boldsymbol{h}}_{\boldsymbol{2}}$(${\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}},{\boldsymbol{x}}_{\boldsymbol{3}}\boldsymbol{)}\mathrm{\le }\mathrm{\ }$$\boldsymbol{0}$$\mathrm{\to }\mathrm{\ \ \ \ \ \ }\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{3}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{12}$$\mathrm{\le }\boldsymbol{0}$ $ \\ $

8) ${\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}},{\boldsymbol{x}}_{\boldsymbol{3}}\mathrm{\ge}$ 0

From Kuhn-Tucker conditions (4) (5) following 4 cases arises

Case 1.

${\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}\boldsymbol{=}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{0}$

From (1), $-\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{4}$= 0

$\boldsymbol{\mathrm{\ }}\mathrm{\therefore }\mathrm{\ }{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{=}\boldsymbol{2}$

From (2), $-\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{+}\boldsymbol{6}$= 0

$\mathrm{\therefore }\mathrm{\ }{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{3}$

Substitute ${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{\ }\boldsymbol{and}\boldsymbol{\ }{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{\ }\boldsymbol{in}\boldsymbol{\ (}\boldsymbol{6}\boldsymbol{)}$

LHS =$\boldsymbol{\ }\boldsymbol{2}\boldsymbol{+}\boldsymbol{3}\boldsymbol{-}\boldsymbol{2}\boldsymbol{=}\boldsymbol{3}\boldsymbol{\ }\boldsymbol{\nleq }\boldsymbol{0}$

Values $x_{1}, x_{2}, x_{3}, \lambda_{1}, \lambda_{2}$ do not satisfies Kuhn-Tucker conditions

$\mathrm{\therefore }\mathrm{We\ Reject\ this\ case}$

Case 2.

${\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}\boldsymbol{\neq }\boldsymbol{0}\boldsymbol{,\ \ }{\boldsymbol{\mathrm{\lambda }}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{0}$

From (4), $x_{1} + x_{2} - 2 = 0$

$\boldsymbol{\mathrm{\ }}\mathrm{\therefore }\mathrm{\ }{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{2} \mathrm{\to } ......(9)$

From (1), $-\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{4}$$\boldsymbol{-}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}$= 0 $\mathrm{\therefore }\mathrm{\ }2{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}= 4 \mathrm{\to }\mathrm{\ }......(10)$ $ \\ $

From (2), $-\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{+}\boldsymbol{6}\boldsymbol{-}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}$= 0

$\mathrm{\therefore }\mathrm{\ }2{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{+}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}= 6 \mathrm{\to }\mathrm{\ }....(11)$

Solving (9) (10) and (11) simultaneously, we get

${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{=}\boldsymbol{0}.\boldsymbol{5}\boldsymbol{;} {\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{1}.\boldsymbol{5}; {\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}\boldsymbol{=}\boldsymbol{3}$

Substitute ${\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}$ in (7)

LHS = $\boldsymbol{2}\left(\boldsymbol{0}.\boldsymbol{5}\right)\boldsymbol{+}\boldsymbol{3}\left(\boldsymbol{1}.\boldsymbol{5}\right)\boldsymbol{-}\boldsymbol{12}\boldsymbol{=-}\boldsymbol{6}.\boldsymbol{5}$$\mathrm{\le }\mathrm{0}$ $ \\ $

Values ${\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{\&\ }{\boldsymbol{x}}_{\boldsymbol{3}}$${,\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}\boldsymbol{\&\ }{\boldsymbol{\mathrm{\lambda }}}_{\boldsymbol{2}}$ $ \\ $

satisfies Kuhn-Tucker conditions (6) (7) $\&$ (8)

Stationary Point ${\boldsymbol{X}}_{\boldsymbol{0}}$ = $\boldsymbol{(}\boldsymbol{0}.\boldsymbol{5},\boldsymbol{1}.\boldsymbol{5},\boldsymbol{3}\boldsymbol{)}$

Since ${,\boldsymbol{\mathrm{\lambda }}}_{\boldsymbol{1}}\boldsymbol{\gt}\boldsymbol{0}$

Z is maximum

$\mathrm{\therefore } {\boldsymbol{z}}_{\boldsymbol{max}}\boldsymbol{=\ }\ \ {-0.5}^2-{1.5}^2-\ 0^2+4\left(0.5\right)+6\left(1.5\right)$ $ = 8.5$

Case 3. ${\boldsymbol{\mathrm{\lambda }}}_{\boldsymbol{1}}\boldsymbol{=}\boldsymbol{0}\boldsymbol{,\ \ }{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}\boldsymbol{\neq }\boldsymbol{0}$

From (5), $\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{3}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{12}\boldsymbol{\ }\boldsymbol{\mathrm{\ \ \ \ }}\boldsymbol{=}\boldsymbol{0}\boldsymbol{\mathrm{\ \ \ \ }}$

$\boldsymbol{\mathrm{\ }}\mathrm{\therefore }\mathrm{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{3}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{12}\boldsymbol{\ }\boldsymbol{\mathrm{\ \ \ \ }} \mathrm{\to }.......(12)$

From (1), $-\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{4}$$\boldsymbol{-}\boldsymbol{2}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}$= 0 $\mathrm{\therefore }\mathrm{\ }2{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{\mathrm{2}}\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}= 4 \mathrm{\to }\mathrm{\ }.....(13)$ $ \\ $

From (2), $-\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{+}\boldsymbol{6}\boldsymbol{-}{\boldsymbol{\mathrm{3}}\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}$= 0

$\mathrm{\therefore }\mathrm{\ }2{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{+}{\boldsymbol{\mathrm{3}}\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}= 6 \mathrm{\to }\mathrm{\ }....(14)$

Solving (12) (13) and (14) simultaneously, we get

${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{=}\frac{\boldsymbol{24}}{\boldsymbol{13}}\boldsymbol{;} {\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\frac{\boldsymbol{36}}{\boldsymbol{13}}; {\boldsymbol{\mathrm{\lambda }}}_{\boldsymbol{2}}\boldsymbol{=}\frac{\boldsymbol{2}}{\boldsymbol{13}}$

substitute ${\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}$in (6)

LHS = $\frac{\boldsymbol{24}}{\boldsymbol{13}}\boldsymbol{+}\frac{\boldsymbol{36}}{\boldsymbol{13}}\boldsymbol{-}\boldsymbol{2}\boldsymbol{=}\frac{\boldsymbol{34}}{\boldsymbol{13}}$$\boldsymbol{\nleq }\boldsymbol{0}$ $\\ $

Values ${\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{\&\ }{\boldsymbol{x}}_{\boldsymbol{3}}$ $ \\ $

do not satisfies Kuhn-Tucker conditions

$\mathrm{\therefore }\mathrm{We\ Reject\ this\ case}$

Case 4. ${\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}\boldsymbol{\neq }\boldsymbol{0}\boldsymbol{,\ \ }{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{2}}\boldsymbol{\neq }\boldsymbol{0}$

From (4), ${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{2}$= 0

$\boldsymbol{\mathrm{\ }}\mathrm{\therefore }\mathrm{\ }{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{2} \mathrm{\to }.....(15)$ $ \\ $

From (5), $\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{3}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{12}\boldsymbol{\ }\boldsymbol{\mathrm{\ \ \ \ }}\boldsymbol{=}\boldsymbol{0}\boldsymbol{\mathrm{\ \ \ \ }}$

$\boldsymbol{\mathrm{\ }}\mathrm{\therefore }\mathrm{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{3}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{12}\boldsymbol{\ }\boldsymbol{\mathrm{\ \ \ \ }} \mathrm{\to }.....(16)$ $ \\ $

Solving (15) and (16) simultaneously, we get

${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{=-}\boldsymbol{6}\boldsymbol{;} {\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{8}; $ $ \\ $

Values ${\boldsymbol{x}}_{\boldsymbol{1}}$ do not satisfies Kuhn-Tucker condition (8) $ \\ $

$\mathrm{\therefore }\mathrm{We\ Reject\ this\ case}$

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

$\mathrm{\therefore }$ ${\boldsymbol{z}}_{\boldsymbol{max}}$ = $8.5$ at $ \\ $

${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{=}\boldsymbol{0}.\boldsymbol{5}\boldsymbol{;}$${\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{1}.\boldsymbol{5}$; ${\boldsymbol{\mathrm{\lambda }}}_{\boldsymbol{1}}\boldsymbol{=}\boldsymbol{3}$