Question

To find the familiar "second derivative test "for a maximum or minimum point of the functions of two variables if${\mathit{f}}_{\mathbf{x}}\mathbf{=}{\mathit{f}}_{\mathbf{y}}\mathbf{=}\mathbf{0}$atlocalid="1664265078344" $\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}$then,

localid="1664265157617" $\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}$ Is maximum point if at $\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}\mathbf{}\mathbf{,}\mathbf{}{\mathit{f}}_{\mathbf{x}\mathbf{x}}\mathbf{>}\mathbf{0}\mathbf{,}{\mathit{f}}_{\mathbf{y}\mathbf{y}}\mathbf{>}\mathbf{0}{\mathit{f}}_{\mathbf{x}\mathbf{x}}{\mathit{f}}_{\mathbf{y}\mathbf{y}}\mathbf{>}{\mathit{f}}_{\mathbf{x}\mathbf{y}}^{\mathbf{2}}$.

$\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}$ Is maximum point if at$\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}\mathbf{}\mathbf{,}\mathbf{}{\mathit{f}}_{\mathbf{x}\mathbf{x}}\mathbf{<}\mathbf{0}\mathbf{,}{\mathit{f}}_{\mathbf{y}\mathbf{y}}\mathbf{<}\mathbf{0}{\mathit{f}}_{\mathbf{x}\mathbf{x}}{\mathit{f}}_{\mathbf{y}\mathbf{y}}\mathbf{>}{\mathit{f}}_{\mathbf{x}\mathbf{y}}^{\mathbf{2}}$

$\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}$ Is neither a maximum nor minimum point if ${\mathit{f}}_{\mathbf{x}\mathbf{x}}{\mathit{f}}_{\mathbf{y}\mathbf{y}}\mathbf{<}{\mathit{f}}_{\mathbf{x}\mathbf{y}}^{\mathbf{2}}$.

Short answer

The second derivative test "for a maximum or minimum point of the functions of two variables if $f_x=f_y=0$at $(a,b)$then,

$(a,b)$Is maximum point if at $(a,b),f_{xx}>0,f_{yy}>0f_{xx}{f}_{yy}>f_{xy}^2$.

$(a,b)$Is maximum point if at$(a,b),f_{xx}<0,f_{yy}<0f_{xx}{f}_{yy}>f_{xy}^2$.

$(a,b)$Is neither a maximum nor minimum point if $f_{xx}{f}_{yy}<f_{xy}^2$.

Step-by-step solution

Given data

The given function is $f_x=f_y=0$at $(a,b)$.

Concept of Taylor series

The Taylor series representation is given as $\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{1}}{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{h}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{x}}\mathbf{+}\mathbf{k}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{x}}\mathbf{)}}^{\mathbf{n}}\mathit{f}\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}$$\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{)}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{1}}{\mathbf{n}\mathbf{!}}{\left(h\frac{\partial }{\partial x}+k\frac{\partial }{\partial x}\right)}^{\mathbf{n}}\mathit{f}\mathbf{(}\mathit{a}\mathbf{,}\mathit{b}\mathbf{)}$.

Calculation of the second derivative terms

The second derivative terms on this Taylor series can be written as

$f^{\text{'}\text{'}}(x,y)=A{h}^2+2Bhk+C{k}^2$

Here, assume that$A=f_{xx},B=f_{yy},C=f_{yy}$.

It is need to prove the second derivative test now, write this expression as follows:

$$\begin{gathered} f^{\text{'}\text{'}}(x,y)=A{h}^2+2Bhk+C{k}^2+\frac{B^2{k}^2}{A}-\frac{B^2{k}^2}{A} \\ {f}^{\text{'}\text{'}}(x,y)=A{h}^2+2Bhk+\left(C-\frac{B^2}{A}\right)k^2+\frac{B^2{k}^2}{A} \\ {f}^{\text{'}\text{'}}(x,y)=A\left(h^2+\frac{2Bhk}{A}+\frac{B^2{k}^2}{A}\right)+\left(C-\frac{B^2}{A}\right)k^2 \end{gathered}$$

Now, add the expression with same denominator and solve further as:

$$\begin{gathered} f^{\text{'}\text{'}}(x,y)=A{\left(h+\frac{Bk}{A}\right)}^2+\left(C-\frac{B^2}{A}\right)k^2 \\ {f}^{\text{'}\text{'}}(x,y)=A{\left(h+\frac{Bk}{A}\right)}^2+\left(\frac{CA-B^2}{A}\right)k^2 \end{gathered}$$

Calculation to find the maximum point for the derivative test f

For all $\text{x}\text{,}y$near enough to $a$as follows:

$\begin{array}{rcl}{f}^{\text{'}\text{'}}(x,y)& >& 0\\ 2\frac{f(x,y)-f\left(a\right)}{{(x-a)}^2}& >& 0\\ f(x,y)-f\left(a\right)& >& 0\\ f(x,y)& >& f\left(a\right)\end{array}$

That is, $f$increases.

Now, by the first derivative test $f$has minimum point at $x=a$:

$\begin{array}{rcl}{f}^{\text{'}\text{'}}\left(a\right)& <& 0\\ \frac{2f(x,y)-f\left(a\right)}{\left(\right(x,y)-a)}& <& 0\\ f(x,y)-f\left(a\right)& <& 0\end{array}$

$f(x,y)<f\left(a\right)$

For all $x$near enough to $a$that is $f$decreases.

Now, by the first derivative test $f$has maximum point at $(x,y)=a$.