Q 10 - Calculus

Question

Theorem $12.45$ is inconclusive when the discriminant, $det\left(H_f\right)$, is zero at a stationary point. In Exercises $10-12$ we ask you to illustrate this fact by analyzing three functions of two variables with stationary points at the origin.

Show that the function $f\left(x,y\right)=x^4+y^4$ has a stationary point at the origin. Show that the discriminant $det\left(H_f\left(0,0\right)\right)=0$. Explain why $f$ has an absolute minimum at the origin. 

Step-by-Step Solution

Answer

• As $f_x=4x^3$ and $f_y=4y^3$. Then by putting both equals to zero, we will get $x=0,y=0$. So origin $\left(0,0\right)$ is the stationary point of the given function.
• $f_{xx}(0,0)=0, f_{yy}(0,0)=0 and f_{xy}(0,0)=0$. So that $det\left(H_f\left(0,0\right)\right)=0$.
• The possible value of the function are $f\ge 0$ and minimum value $0$ lies on $\left(0,0\right)$. So $f$ is the absolute minimum at origin.

Step 1. Given Information

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 $\left(0,0\right)$.

Step 2. To find stationary point

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 :-

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

and

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

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.

Step 3. To show d e t H f 0 , 0 = 0

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 :-

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

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 :-

$$\begin{gathered} det\left(H_f\left(0,0\right)\right)=\left(0\times 0\right)-{\left(0\right)}^2 \\ \Rightarrow det\left(H_f\left(0,0\right)\right)=0 \end{gathered}$$

Hence proved.

Step 4. To check absolute minimum

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 $f\left(x,y\right)\ge 0$

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

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

This is the possible minimum value of the function.

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