91Ó°ÊÓ

We have given the following function :-

$f(x,y)=x^4+y^4$.

We have to show that $f(x,y)$has a stationary point at the origin $\left(0,0\right)$.

The given function is :-

$f\left(x,y\right)=x^4+y^4$

Now partially differentiate this function with respect to $x$and $y$, then we have :-

$f_x=4x^3$

and

$f_y=4y^3$

Now put $f_x=0=f_y$, then we have :-

$$\begin{gathered} 4x^3=0 \\ ⇒x^3=0 \\ ⇒x=0 \end{gathered}$$

and

$$\begin{gathered} 4y^3=0 \\ ⇒y^3=0 \\ ⇒y=0 \end{gathered}$$

That is the stationary point is $\left(0,0\right)$.

Hence it is proved that the function$f\left(x,y\right)=x^4+y^4$has stationary point at origin.

We find that :-

$f_x=4x^3$and $f_y=4y^3$

Now find second derivatives :-

$f_{xx}=12x^2$. Then :-

$f_{xx}\left(0,0\right)=0$

and

$f_{yy}=12y^2$. Then :-

role="math" $f_{yy}\left(0,0\right)=0$

and

$f_{xy}=0$

We know that $det\left(H_f\right)$is defined as $det\left(H_f\right)=f_{xx}fyy-{\left(f_{xy}\right)}^2$.

Then :-

role="math" $$\begin{gathered} det\left(H_f\left(0,0\right)\right)=\left(0×0\right)-{\left(0\right)}^2 \\ ⇒det\left(H_f\left(0,0\right)\right)=0 \end{gathered}$$

Hence proved.

The given function is :-

$f\left(x,y\right)=x^4+y^4$

As the powers of both $x$and $y$is $4$. Also relation between $x^4$and $y^4$is addition.

So we can say that the value of the function is always non negative.

That is role="math" $f\left(x,y\right)â‰\yen 0$

At the origin $\left(0,0\right)$the value of the function is :-

$$\begin{gathered} f\left(0,0\right)=0+0 \\ ⇒f\left(0,0\right)=0 \end{gathered}$$

This is the possible minimum value of the function.

So we can conclude that the given function$f$is absolute minimum at origin.