To find the maximum and the minimum points of the given function.

${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{+}\mathbf{10}$

A function is given as $f(x,y)=x^2+y^2+2x-4y+10$.

To evaluate maximum and minimum points of a function$\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}$the following points should be noted:

1. ${\mathit{f}}_{\mathbf{x}}\mathbf{=}\mathbf{0}\mathbf{=}{\mathit{f}}_{\mathbf{y}}$ At $\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}$

2. $\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}$is a minimum point if ${\mathit{f}}_{\mathbf{xx}}\mathbf{>}\mathbf{0}\mathbf{ }\mathbf{,}\mathbf{}{\mathit{f}}_{\mathbf{yy}}\mathbf{>}\mathbf{0}$, and ${\mathit{f}}_{\mathbf{xx}}{\mathit{f}}_{\mathbf{yy}}\mathbf{>}{\mathit{f}}_{\mathbf{xy}}^{\mathbf{2}}$

3. $\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}$is a maximum point if ${\mathit{f}}_{\mathbf{xx}}\mathbf{<}\mathbf{0}\mathbf{ }\mathbf{,}\mathbf{}{\mathit{f}}_{\mathbf{yy}}\mathbf{<}\mathbf{0}$, and${\mathit{f}}_{\mathbf{xx}}{\mathit{f}}_{\mathbf{yy}}\mathbf{>}{\mathit{f}}_{\mathbf{xy}}^{\mathbf{2}}$

4.$\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}$is neither a maximum nor a minimum point if${\mathit{f}}_{\mathbf{xx}}{\mathit{f}}_{\mathbf{yy}}\mathbf{<}{\mathit{f}}_{\mathbf{xy}}^{\mathbf{2}}$that is${\mathit{f}}_{\mathbf{yy}}$and${\mathit{f}}_{\mathbf{xx}}$are of opposite sign.

Here, $f_x,f_y$are the first order partial derivatives and $f_{xx},f_{yy} ,f_{xy}$are the second order partial derivatives.

Consider the given function as follows:

$f(x,y)=x^2+y^2+2x-4y+10$ …(1)

Differentiate (1) partially with respect to $x$as shown below:

$$\begin{gathered} f_x(x,y)=\frac{\partial }{\partial x}\left(x^2+y^2+2x-4y+10\right) \\ {f}_x(x,y)=\frac{\partial }{\partial x}{x}^2+\frac{\partial }{\partial x}{y}^2+\frac{\partial }{\partial x}2x-\frac{\partial }{\partial x}4y+\frac{\partial }{\partial x}10 \end{gathered}$$

$f_x(x,y)=2x+2$ …(2)

Differentiate (1) partially with respect to $y$as shown below:

$$\begin{gathered} f_y(x,y)=\frac{\partial }{\partial y}\left(x^2+y^2+2x-4y+10\right) \\ {f}_y(x,y)=\frac{\partial }{\partial y}{x}^2+\frac{\partial }{\partial y}{y}^2+\frac{\partial }{\partial y}2x-\frac{\partial }{\partial y}4y+\frac{\partial }{\partial y}10 \end{gathered}$$

$f_y(x,y)=2y-4$ …(3)

Equate equation (2) with zero as follow:

$\begin{array}{rcl}{f}_x(x,y)& =& 2x+2\\ 0& =& 2x+2\\ x& =& -1\end{array}$

Now, equate equation (3) with zero as follows:

$\begin{array}{rcl}{f}_y(x,y)& =& 2y-4\\ 0& =& 2y-4\\ y& =& 2\\ & & \end{array}$

The obtained point is $(-1,2)$.

Now, differentiate (2) and (3) again as shown below:

$f_{xx}(x,y)=2$ …(4)

$f_{yy}(x,y)=2$ …(5)

$f_{xy}(x,y)=0$ …(6)

Now, use the value obtained in equation $\left(4\right),\left(5\right)$and (6) to check the maximum and minimum point.

At $(-1,2)$,$f_{xx}>0 ,f_{yy}>0$and $f_{xx}{f}_{yy}>{f_{xy}}^2$.

Therefore, the point $(-1,2)$is minimum point of the given function.