Subject to 2x${}_{1}$-5x${}_{2}$ $\boldsymbol{\mathrm{\le}}$ 98

X${}_{1,}$X${}_{2}$ $\boldsymbol{\mathrm{\ge}}$ 0.

Let $$f({\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{)} = \boldsymbol{2}{\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{1}}\boldsymbol{-}{\boldsymbol{7}\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{2}}\boldsymbol{\mathrm{+}}\boldsymbol{\mathrm{12}}\boldsymbol{\mathrm{\ }}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{\mathrm{\ }}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{\mathrm{\ }}\boldsymbol{;}\boldsymbol{h}= $\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{5}\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{98}\boldsymbol{;}\boldsymbol{\ }$$

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

Lagrangian function L = f - $\lambda$h

$$L = \boldsymbol{(}\boldsymbol{2}{\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{1}}\boldsymbol{-}{\boldsymbol{7}\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{2}}\boldsymbol{\mathrm{+}}\boldsymbol{\mathrm{12}}\boldsymbol{\mathrm{\ }}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{\mathrm{\ }}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{\mathrm{\ }}\boldsymbol{)}-\boldsymbol{\lambda}(\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{5}\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{98}\boldsymbol{)}$$

$$L = \boldsymbol{2}{\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{1}}\boldsymbol{-}{\boldsymbol{7}\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{2}}\boldsymbol{\mathrm{+}}\boldsymbol{\mathrm{12}}\boldsymbol{\mathrm{\ }}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{\mathrm{\ }}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{\mathrm{\ }}-\boldsymbol{\lambda}\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{-}\boldsymbol{\mathrm{\ }}\boldsymbol{\mathrm{\lambda}}{\boldsymbol{5}\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{+}\boldsymbol{\mathrm{\ }}\boldsymbol{\mathrm{\lambda}}\boldsymbol{98}$$

Kuhn-Tucker conditions are:

$$1) \frac{\partial L}{\partial x_1} = 0 \mathrm{\to}\boldsymbol{4}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{12}\boldsymbol{\mathrm{\ }}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{2}\boldsymbol{\mathrm{\lambda}}\boldsymbol{=}\boldsymbol{0}$$

$$2) \frac{\partial L}{\partial x_2} = 0 \mathrm{\to}\boldsymbol{12}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{\mathrm{\ }}{\boldsymbol{-}\boldsymbol{14}\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{5}\boldsymbol{\mathrm{\lambda}}\boldsymbol{=}\boldsymbol{0}$$

$$3) \boldsymbol{\mathrm{\lambda}}\boldsymbol{h}({\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{)} \boldsymbol{=}\boldsymbol{0}\mathrm{\to}{\boldsymbol{\mathrm{\lambda}}}_{\boldsymbol{1}}(\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{5}\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{98}\boldsymbol{)\ }\boldsymbol{=}\boldsymbol{0}$$

$$4) \boldsymbol{h}({\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{)}\mathrm{\le }\mathrm{\ }\boldsymbol{0}\mathrm{\to }\mathrm{\ \ \ \ \ \ }\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{5}\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{98}\mathrm{\le }\boldsymbol{0}$$

$$5) {\boldsymbol{h}}_{\boldsymbol{2}}({\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{)}\mathrm{\le }\mathrm{\ }\boldsymbol{0}\mathrm{\to }\mathrm{\ \ \ \ \ \ }\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{5}\mathrm{\le }\boldsymbol{0}$$

$$6) \boldsymbol{\mathrm{\lambda}},{\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}\mathrm{\ge} 0$$

From Kuhn-Tucker conditions (3) following 2 cases arises

Case 1. $$\boldsymbol{\mathrm{\lambda}}\boldsymbol{=}\boldsymbol{0}$$

From (1), $$\boldsymbol{4}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{12}\boldsymbol{\mathrm{\ }}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{0}\boldsymbol{=}\boldsymbol{0}$$ $$\boldsymbol{\mathrm{\ }}\mathrm{\therefore }\mathrm{\ }\boldsymbol{4}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{12}\boldsymbol{\mathrm{\ }}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{0}$$

From (2), $$\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{\mathrm{\ }}{\boldsymbol{-}\boldsymbol{14}\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{0}= 0$$ $$\mathrm{\therefore }\mathrm{\ }\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{\mathrm{\ }}{\boldsymbol{-}\boldsymbol{14}\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{0}$$

Solving simultaneously $${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{=}\boldsymbol{0}\boldsymbol{\ \ \ \ }\boldsymbol{;}\boldsymbol{\ \ }{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{0}$$

substitute ${\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{x}}_{\boldsymbol{2}}$ in (4)

$$LHS= \boldsymbol{2}\left(\boldsymbol{0}\right)\boldsymbol{+}\boldsymbol{5}\left(\boldsymbol{0}\right)\boldsymbol{-}\boldsymbol{98}\boldsymbol{=-}\boldsymbol{98}$$

Values ${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{\&\ }{\boldsymbol{x}}_{\boldsymbol{2}}$satisfies Kuhn-Tucker conditions

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

Z is maximum $$\mathrm{\therefore } {\boldsymbol{z}}_{\boldsymbol{max}}\boldsymbol{=\ }{\ 2({0)}^2-7(0)}^2+12(0)(0)$$ $$ = 0$$

Case 2. $$\boldsymbol{\mathrm{\lambda}}\boldsymbol{\mathrm{\neq }}\boldsymbol{0}$$

From (3), $$\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{5}\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{98}\boldsymbol{=}\boldsymbol{0}$$

$$\boldsymbol{\mathrm{\ }}\mathrm{\therefore }\mathrm{\ }\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{5}\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{98} \mathrm{\to }\mathrm{\ }.....(6) $$

Solving (1) (2) & (6) simultaneously, we get $${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{=}\boldsymbol{44}\boldsymbol{\ }\boldsymbol{;} {\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{2}\boldsymbol{;} \boldsymbol{\mathrm{\lambda}}\boldsymbol{\mathrm{=}}100;$$

Values ${\boldsymbol{x}}_{\boldsymbol{1}},{\boldsymbol{\&\ }\boldsymbol{x}}_{\boldsymbol{2}}$satisfies Kuhn-Tucker conditions

Stationary Point ${\boldsymbol{X}}_{\boldsymbol{0}}$= $\boldsymbol{(}\boldsymbol{44},\boldsymbol{2}\boldsymbol{)}$

Z is maximum $$\mathrm{\therefore } {\boldsymbol{z}}_{\boldsymbol{max}}\boldsymbol{=\ }\ \ {\ 2({44)}^2-7(2)}^2+12(44)(2)$$ $$ = 4900$$

Hence from Case 1 & 2 We conclude that

$\mathrm{\therefore }$ ${\boldsymbol{z}}_{\boldsymbol{max}}$ = $4900$ at ${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{=}\boldsymbol{44}\boldsymbol{;}$${\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{2}$; $\boldsymbol{\mathrm{\lambda}}\boldsymbol{\mathrm{=}}$100;