Let $$f(x_1,x_2) = 4x_1+8x_2-x^2_1- x^2_2\ ; h = x_1+x_2-4$$ and $\lambda$ be Lagrangian multiplier

Lagrangian function L = f - $\lambda$h

L = $(x_1+8x_2$ $-$ $x^2_1$ $-$ $x^2_2)$ $--$ $\lambda$($x_1+x_2-4)$

L = $x_1+8x_2$ $-$ $x^2_1$ $-$ $x^2_2$ $-$ $\lambda$$x_1-\mathrm{\lambda}x_2+4\mathrm{\lambda}$ Conditions to find stationary points are $\frac{\partial L}{\partial x_1}$ = 0 $\mathrm{\to}$ $4-\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{-}\boldsymbol{\mathrm{\lambda}}\boldsymbol{=}\boldsymbol{0}\mathrm{\therefore }$ $2{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}\boldsymbol{\mathrm{\lambda}}\boldsymbol{=}\boldsymbol{4}\mathrm{\to}$ (1)

$\frac{\partial L}{\partial x_2}$ = 0 $\mathrm{\to}$ $8-\boldsymbol{2}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{-}\boldsymbol{\mathrm{\lambda}}\boldsymbol{=}\boldsymbol{0}$$\mathrm{\therefore }$ $2{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{+}\boldsymbol{\mathrm{\lambda}}\boldsymbol{=}\boldsymbol{8}\mathrm{\to}$(2)

$\frac{\partial L}{\partial \boldsymbol{\mathrm{\lambda}}}$ = 0 $\mathrm{\to}$ $\boldsymbol{-}{\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{-}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{+}\boldsymbol{4}\boldsymbol{=}\boldsymbol{0}\mathrm{\therefore }$ ${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{+}{\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{4}\mathrm{\to}$(3)

Solving (1) (2) and (3) simultaneously $${\boldsymbol{x}}_{\boldsymbol{1}}\boldsymbol{=}\boldsymbol{1}\boldsymbol{\ }\boldsymbol{;} {\boldsymbol{x}}_{\boldsymbol{2}}\boldsymbol{=}\boldsymbol{3}\boldsymbol{;} \boldsymbol{\mathrm{\lambda}}\boldsymbol{\mathrm{=}}2;$$ Stationary Point ${\boldsymbol{X}}_{\boldsymbol{0}} = (1,3)$

Now, $\boldsymbol{\ \ \ }\frac{{\partial }^2L}{\partial {\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{1}}\boldsymbol{\mathrm{\ }}}=-2$; $\frac{{\partial }^2L}{\partial x_{1\ }\partial x_2\boldsymbol{\mathrm{\ }}}=0;$ $\frac{{\partial }^2L}{\partial {\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{2}}\boldsymbol{\mathrm{\ }}}=-2$ ; $\frac{\partial h}{\partial x_1}=1;$ $\frac{\partial h}{\partial x_2}=1$

$\boldsymbol{\Delta}$ = $\left| \begin{array}{ccc} 0 & \frac{\partial h}{\partial x_1} & \frac{\partial h}{\partial x_2} \\ \frac{\partial h}{\partial x_1} & \frac{{\partial }^2L}{\partial {\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{1}}\boldsymbol{\mathrm{\ }}} & \frac{{\partial }^2L}{\partial x_{1\ }\partial x_2\boldsymbol{\mathrm{\ }}} \\ \frac{\partial h}{\partial x_2} & \frac{{\partial }^2L}{\partial x_{1\ }\partial x_2\boldsymbol{\mathrm{\ }}} & \frac{{\partial }^2L}{\partial {\boldsymbol{x}}^{\boldsymbol{2}}_{\boldsymbol{2}}\boldsymbol{\mathrm{\ }}} \end{array} \right|$

$ = \left| \begin{array}{ccc} 0 & 1 & 1 \\ 1 & -2 & 0 \\ 1 & 0 & -2 \end{array} \right| = 0-1\left(-2-0\right)+1\left(0+2\right) = 2 + 2$

= 4 $\gt$ 0 Since $\boldsymbol{\Delta}$ is positive Z is maximum at ${\boldsymbol{X}}_{\boldsymbol{0}} = (1,3)$ $$\mathrm{\therefore } {\boldsymbol{z}}_{\boldsymbol{max}}\boldsymbol{=\ }\ {4\left(1\right)+8\left(3\right)-\ 1^2-3}^2$$ $$ = 4+24-\ 1-9$$ $$= 18$$