For two variables i.e $\quad u=f(x,y)$:
Condition for Maxima and Minima is given by: $$\frac{\partial f}{\partial x}=0 \quad , \quad \frac{\partial f}{\partial y}=0 \quad \implies \text{Critical points}$$ $$H(x,y)= f_{xx}f_{yy} \, - \, (fxy)^2 \quad \implies \text{Checking function}$$ $$\text{For Maxima:} \quad f_{xx}<0 \, , \, H(x,y)>0 \quad |_{\text{(x,y)=Critical points}}$$ $$\text{For Minima:} \quad f_{xx}>0 \, , \, H(x,y)>0 \quad |_{\text{(x,y)=Critical points}}$$
My Thoughts:
For one variable i.e $\quad u=f(x)$
Condition for Maxima and Minima is given by:
$$\frac{d f}{d x}=0 \quad \quad \implies \text{Critical points}$$
$$\text{For Maxima:} \quad f_{x}<0 \, \quad |_{\text{x=Critical points}}$$
$$\text{For Minima:} \quad f_{xx}>0 \, \quad |_{\text{x=Critical points}}$$
Extending this concept:
$$du= \frac{\partial f}{\partial x}dx \, + \, \frac{\partial f}{\partial y}dy \quad [\because u=f(x,y)]$$
For Critical Points: $\quad du=0$
$$\implies \frac{\partial f}{\partial x}dx \, + \, \frac{\partial f}{\partial y}dy =0 $$
$$\implies \frac{\partial f}{\partial x} =0 \quad \& \quad \frac{\partial f}{\partial y} =0 \quad \dots(1)$$
For Maxima : $\quad d^2u <0 \, $ &
For Minima : $\quad d^2u >0 \, $
$$\therefore d^2u=d(du)$$
$$=d(\frac{\partial f}{\partial x}dx \, + \, \frac{\partial f}{\partial y}dy)$$
$$\implies d^2u=d(\frac{\partial f}{\partial x}dx)\, + \, d(\frac{\partial f}{\partial y}dy) \quad \dots (2)$$
let $M(x,y)=\frac{\partial f}{\partial x}dx $, then
$$d(M)=\frac{\partial M}{\partial x}dx \, + \, \frac{\partial M}{\partial y}dy \quad \dots (3)$$
$$\frac{\partial M}{\partial x}=\frac{\partial }{\partial x} (\frac{\partial f}{\partial x}dx)$$
$$=\frac{\partial ^2 f}{\partial x^2}dx \, + \, \frac{\partial f}{\partial x}\frac{d }{d x} dx $$
$$\implies \frac{\partial M}{\partial x}=\frac{\partial ^2 f}{\partial x^2}dx \, + \, \frac{\partial f}{\partial x}\frac{d^2 x }{d x} \quad \dots (4) $$
$$\text{similarly,} \quad \frac{\partial M}{\partial y}=\frac{\partial ^2 f}{\partial y \, \partial x}dx \, + \, \frac{\partial f}{\partial x}\frac{d }{d y} dx $$
$$=\frac{\partial ^2 f}{\partial y \, \partial x}dx \, + \, \frac{\partial f}{\partial x}\cdot 0 $$
$$\implies \frac{\partial M}{\partial y}=\frac{\partial ^2 f}{\partial y \, \partial x}dx \quad \dots (5) $$
$$\implies dM= (\frac{\partial ^2 f}{\partial x^2}dx \, + \, \frac{\partial f}{\partial x}\frac{d^2 x }{d x})dx \, + \, (\frac{\partial ^2 f}{\partial y \, \partial x}dx)dy $$
$$\implies dM=\frac{\partial ^2 f}{\partial x^2}(dx)^2 \, + \, \frac{\partial f}{\partial x}d^2 x \, + \, \frac{\partial ^2 f}{\partial y \, \partial x}dxdy \quad \dots (6)$$
similarly let $N(x,y)=\frac{\partial f}{\partial y}dy $, then
$$d(N)=\frac{\partial N}{\partial x}dx \, + \, \frac{\partial N}{\partial y}dy \quad \dots (7)$$
$$\frac{\partial N}{\partial x}=\frac{\partial }{\partial x} (\frac{\partial f}{\partial y}dy)$$
$$=\frac{\partial ^2 f}{\partial x \, \partial y}dy $$
$$\text{and,} \quad \frac{\partial N}{\partial y}=\frac{\partial ^2 f}{\partial y^2}dy \, + \, \frac{\partial f}{\partial y}\frac{d^2 y }{d y} $$
$$\implies dN=\frac{\partial ^2 f}{\partial x \, \partial y}dy\, dx \, + \, \frac{\partial ^2 f}{\partial y^2}(dy)^2 \, + \, \frac{\partial f}{\partial y}d^2 y \, + \, \quad \dots (8)$$
$$\implies d^2 u=\frac{\partial ^2 f}{\partial x^2}(dx)^2 +\frac{\partial f}{\partial x}d^2 x +\frac{\partial ^2 f}{\partial y \partial x}dx \, dy +\frac{\partial ^2 f}{\partial x \partial y}dy \, dx \, + \frac{\partial ^2 f}{\partial y^2}(dy)^2 +\frac{\partial f}{\partial y}d^2 y \quad \dots (9)$$
let us assume $f(x,y)$ to be a continuous function
,then according to Schwarz's Theorem: $$\frac{\partial ^2 f}{\partial y \partial x}=\frac{\partial ^2 f}{\partial x \partial y}$$
$$\therefore d^2u=\frac{\partial ^2 f}{\partial x^2}(dx)^2 +\frac{\partial f}{\partial x}d^2 x +2 \cdot \frac{\partial ^2 f}{\partial y \partial x}dx \, dy \, + \frac{\partial ^2 f}{\partial y^2}(dy)^2 +\frac{\partial f}{\partial y}d^2 y \quad \dots (10) $$
so, for maxima:
$$\frac{\partial ^2 f}{\partial x^2}(dx)^2 +\frac{\partial f}{\partial x}d^2 x +2 \cdot \frac{\partial ^2 f}{\partial y \partial x}dx \, dy \, + \frac{\partial ^2 f}{\partial y^2}(dy)^2 +\frac{\partial f}{\partial y}d^2 y <0 \quad |_{\text{(x,y)=Critical points}}$$
$$\implies \frac{\partial ^2 f}{\partial x^2}(dx)^2 + \frac{\partial ^2 f}{\partial y^2}(dy)^2 + 2 \frac{\partial ^2 f}{\partial y \partial x}dx \, dy <0 \quad |_{\text{(x,y)=Critical points}}$$
$$[\because \frac{\partial f}{\partial x}=0 \quad , \quad \frac{\partial f}{\partial y}=0 \quad |_{(x,y)= \text{Critical points}}]$$
$$\implies (f_{xx})\cdot(dx)^2 + (f_{yy})\cdot(dy)^2 + 2 (f_{xy})\cdot dx \, dy <0 \,|_{\text{(x,y)=Critical points}}$$
similarly, for minima:
$$ \frac{\partial ^2 f}{\partial x^2}(dx)^2 + \frac{\partial ^2 f}{\partial y^2}(dy)^2 + 2 \frac{\partial ^2 f}{\partial y \partial x}dx \, dy >0 \quad |_{\text{(x,y)=Critical points}}$$
$$\implies (f_{xx})\cdot(dx)^2 + (f_{yy})\cdot(dy)^2 + 2 (f_{xy})\cdot dx \, dy >0 \,|_{\text{(x,y)=Critical points}}$$
Now how to proceed to derive the condition,please help...