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

Marks: 8M

Year: Dec 2015

Let$$ f(x_1,x_2) = 2x_1+3x_2-x^2_1-2x^2_2;h_1= x_1+3x_2-6;\ h_2= 5x_1+{2x}_2-10$$

Let $\lambda$ be Lagrangian multiplier

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

L = $(\mathrm{\ }2x_1+3x_2-x^2_1$ $-$ $2x^2_2)$ $-$ ${\mathrm{\lambda }}_1$ ( $x_1+3x_2-6)\ -$ ${\mathrm{\lambda}}_2$( $\mathrm{\ }5x_1+{2x}_2-10\mathrm{\ })$

L = $\mathrm{\ \ \ }\ 2x_1+3x_2-x^2_1\mathrm{\ }-\mathrm{\ }2x^2_2\mathrm{\ }--\mathrm{\ }{\mathrm{\lambda}}_1\mathrm{\ }x_1-{\mathrm{\lambda}}_13x_2+{\mathrm{\lambda}}_16\ --\mathrm{\ }{\mathrm{\lambda }}_25x_1-\mathrm{\ }{\mathrm{\lambda}}_2{2x}_2+\mathrm{\ }{\mathrm{\lambda}}_210\mathrm{\ }$

Kuhn-Tucker conditions are:

1) $\frac{\partial L}{\partial x_1}$ = 0 $\mathrm{\to}$ $2-2x_1-{\mathrm{\lambda}}_1-5{\mathrm{\lambda}}_2=0$

2) $\frac{\partial L}{\partial x_2}$ = 0 $\mathrm{\to}$ $\mathrm{3}-4x_2-{\mathrm{3}\mathrm{\lambda}}_1-2{\mathrm{\lambda}}_2=0$

3) ${\mathrm{\lambda}}_1h$ ($x_1,x_2)$ $=0$ $\mathrm{\to}$ ${\mathrm{\lambda}}_1$ ($x_1+3x_2-6)\ $ $=0$

4) ${\mathrm{\lambda}}_2h$ ($x_1,x_2)$ $=0\mathrm{\ \ \ }\mathrm{\to }\mathrm{\ \ \ }{\mathrm{\lambda}}_2\mathrm{\ (}{5x}_1+{2x}_2-10\mathrm{\ })\ \mathrm{\ \ \ \ }=0\mathrm{\ \ \ \ }$

5) $h_1$($x_1,x_2)\mathrm{\le }\mathrm{\ }$ $0$ $\mathrm{\to }\mathrm{\ \ \ \ \ \ }x_1+3x_2-6$ $\mathrm{\le }0$ 6) $h_2$($x_1,x_2)\mathrm{\le }\mathrm{\ }$ $0$ $\mathrm{\to }\mathrm{\ \ \ \ \ }x_1+{2x}_2-10\mathrm{\ }$ $\mathrm{\le }0$ 7) ${\mathrm{\lambda}}_1,{\mathrm{\lambda}}_2,x_1,x_2$ $\mathrm{\ge}$ 0

From Kuhn-Tucker conditions (3) (4) following 4 cases arises

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

From (1), $2-2x_1-0-0$ = 0 $$\mathrm{\ }\mathrm{\therefore }\mathrm{\ }x_1=1$$ From (2), $\mathrm{3}-4x_2-0-0$= 0 $$\mathrm{\therefore }\mathrm{\ }x_2=\frac{3}{4}$$ Substitute $x_1\ and\ x_2\ in\ (5)$

LHS =$\ 1+3\ \left(\frac{3}{4}\right)-6=-\frac{11}{4}\lt0$

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

LHS =$\ 1+2\left(\frac{3}{4}\right)-10=-\frac{15}{2}\lt0$

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

Stationary Point $X_0$ = $(1,\frac{3}{4})$

Z is maximum $$\mathrm{\therefore } z_{max}=\ \ \ {2\left(1\right)+3\left(\frac{3}{4}\right)-\ 1^2-2\left(\frac{3}{4}\right)}^2$$ $$ = \frac{17}{8}$$

Case 2. $${\mathrm{\lambda}}_1\neq 0,\ \ {\mathrm{\lambda}}_2=0$$

From (3), $x_1+3x_2-6$= 0

$$\mathrm{\ }\mathrm{\therefore }\mathrm{\ }x_1+3x_2=6 \mathrm{\to }.......(8)$$

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

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

From (2), $\mathrm{3}-4x_2-{\mathrm{3}\mathrm{\lambda}}_1-0$= 0

$$\mathrm{\therefore }\mathrm{\ }4x_2+3{\mathrm{\lambda}}_1= 3 \mathrm{\to }\mathrm{\ }......(10)$$

Solving (8) (9) and (10) simultaneously, we get $$x_1=\frac{3}{2}; x_2=\frac{3}{2}; {\mathrm{\lambda}}_1=-1$$

Values ${\mathrm{\lambda}}_1$ does not satisfies Kuhn-Tucker conditions (7)

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

Case 3. $${\mathrm{\lambda}}_1=0,\ \ {\mathrm{\lambda}}_2\neq 0$$

From (4), 5$x_1+{2x}_2-10\mathrm{\ }\ \mathrm{\ \ \ \ }=0\mathrm{\ \ \ \ }$

$$\mathrm{\ }\mathrm{\therefore }{5x}_1+{2x}_2=10\ \mathrm{\ \ \ \ } \mathrm{\to }.........(11)$$

From (1), $2-2x_1-0-5{\mathrm{\lambda}}_2$= 0

$$\mathrm{\therefore }\mathrm{\ }2x_1+5{\mathrm{\lambda}}_2= 2 \mathrm{\to }\mathrm{\ }..........(12)$$

From (2), $\mathrm{3}-4x_2-0-2{\mathrm{\lambda}}_2$= 0

$$\mathrm{\therefore }4x_2+2{\mathrm{\lambda}}_2= 3 \mathrm{\to }\mathrm{\ }......(13)$$

Solving (11) (12) and (13) simultaneously, we get $$x_1=\frac{89}{54}; x_2=\frac{95}{108}; {\mathrm{\lambda}}_2=-\frac{7}{27}$$

Values ${\mathrm{\lambda}}_2$ does not satisfies Kuhn-Tucker conditions (7)

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

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

From (3), $x_1+3x_2-6\ $= 0

$$\mathrm{\ }\mathrm{\therefore }\mathrm{\ }x_1+3x_2=6 \mathrm{\to }...........(14)$$

From (4), 5$x_1+{2x}_2-10\mathrm{\ \ \ \ \ }=0\mathrm{\ \ \ \ }$

$$\mathrm{\ }\mathrm{\therefore }\mathrm{5}x_1+{2x}_2=10\mathrm{\ \ }\ \mathrm{\ \ \ \ } \mathrm{\to }........(15)$$

Solving (14) and (15) simultaneously, we get $$x_1=\frac{18}{13}; x_2=\frac{20}{13}; $$

From (1), $2-2x_1-{\mathrm{\lambda}}_1-5{\mathrm{\lambda}}_2=0$
$$\mathrm{\therefore }\mathrm{\ }2-2(\frac{18}{13})-{\mathrm{\lambda}}_1-5{\mathrm{\lambda}}_2= 0$$

$${\mathrm{\lambda}}_1+5{\mathrm{\lambda}}_2=-\frac{10}{13} \mathrm{\to }\mathrm{\ }..........(16)$$

From (2), $\mathrm{3}-4x_2-{\mathrm{3}\mathrm{\lambda}}_1-2{\mathrm{\lambda}}_2$= 0 $$\mathrm{\therefore }\mathrm{3}-4(\frac{20}{13})-{\mathrm{3}\mathrm{\lambda}}_1-2{\mathrm{\lambda}}_2= 0 $$

$${\mathrm{3}\mathrm{\lambda}}_1+2{\mathrm{\lambda }}_2\mathrm{=-}\frac{\mathrm{41}}{\mathrm{13}}\mathrm{\ }\mathrm{\to }\mathrm{\ }.........(17)$$

Solving (16) & (17) we get ${\mathrm{\lambda}}_1$ $\mathrm{=-}\frac{\mathrm{185}}{\mathrm{169}}\mathrm{\ }$ ${\mathrm{\lambda}}_2\mathrm{=}\frac{\mathrm{11}}{\mathrm{169}}$

Value of ${\mathrm{\lambda}}_1$ do not satisfy Kuhn-Tucker conditions

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

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

$\mathrm{\therefore }$ $z_{max}$ = $\frac{17}{8}$ at Stationary Point $X_0$ = $(1,\frac{3}{4})$