Q 11 - Calculus

Question

Theorem $12.45$ is inconclusive when the discriminant, $det\left(H_f\left(0,0\right)\right)$, 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 $g\left(x,y\right)=-\left(x^4+y^4\right)$ has a stationary point at the origin. Show that the discriminant $det\left(H_g(0,0)\right)=0$. Explain why $g$ has an absolute maximum at the origin.

Step-by-Step Solution

Answer

• As $g_x=-4x^3$ and $g_y=-4y^3$. 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.
• $g_{xx}(0,0)=0, g_{yy}(0,0)=0 and g_{xy}(0,0)=0$. So that $det\left(H_g\left(0,0\right)\right)=0$.
• The possible value of the function are $g\le 0$ and maximum value $0$ lies on $\left(0,0\right)$. So $g$ is the absolute maximum at origin.

Step 1. Given Information

We have given the following function :-

$g\left(x,y\right)=-\left(x^4+y^4\right)$.

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

Step 2. To find stationary point

The given function is :-

$g\left(x,y\right)=-\left(x^4+y^4\right)$.

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

$g_x=-4x^3$

and

$g_y=-4y^3$.

Now put $g_x=0=g_y$, 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 $g\left(x,y\right)=-\left(x^4+y^4\right)$ has stationary point at origin. 

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

We find that :-

$g_x=-4x^3$ and $g_y=-4y^3$.

Now find second derivatives :-

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

$g_{xx}(0,0)=0$

and

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

$g_{yy}(0,0)=0$

and

$g_{xy}=0$

We know that $det\left(H_g\right)$ is defined as $det\left(H_g\right)=g_{xx}{g}_{yy}-{\left(g_{xy}\right)}^2$.

Then :-

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

Hence proved.

Step 4. To check absolute minimum

The given function is :- 

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

We know that :-

$$\begin{gathered} x^4,y^4\ge 0 \\ \Rightarrow x^4+y^4\ge 0 \\ \Rightarrow -\left(x^4+y^4\right)\le 0 \end{gathered}$$

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

That is $g\left(x,y\right)\le 0$.

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

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

This is the possible maximum value of the function.

So we can conclude that the given function $g$ is absolute maximum at origin.