Extrema of Functions of Two Variables

[harpazo]

Find the critical points and test for relative extrema. List the critical points for which the Second Partials Test fails.

Given: f (x, y) = (x - 1)^2 (y + 4)^2

I found the partial derivative for x and y to be the following:

f_x = 2 (x - 1)(y + y)^2

f_y = 2 (y + 4)(x - 1)^2

I solved for x and y by setting f_x and f_y to 0.

I then found the critical points to be (1, 0) and (0, -4).

I evaluated f (x, y) at each critical point to find the relative minima. My answer for relative minima is the point
in space (0, -4, 0).

I found that the Second Partials Test fails.

The textbook's answer reveals that the Second Partials Test fails in this case. However, the textbook also reveals that there is no relative extrema but rather absolute minima at (1, a, 0) and (b, -4, 0).

1. Why is the answer an ABSOLUTE MINIMA and not a RELATIVE EXTREMA?

2. In Terms of the absolute MINIMA, where do the letters a and b come from or can it simply be a textbook typo?

 

[I like Serena]

Find the critical points and test for relative extrema. List the critical points for which the Second Partials Test fails.

Given: f (x, y) = (x - 1)^2 (y + 4)^2

I found the partial derivative for x and y to be the following:

f_x = 2 (x - 1)(y + y)^2

f_y = 2 (y + 4)(x - 1)^2

I solved for x and y by setting f_x and f_y to 0.

I then found the critical points to be (1, 0) and (0, -4).

I evaluated f (x, y) at each critical point to find the relative minima. My answer for relative minima is the point
in space (0, -4, 0).

I found that the Second Partials Test fails.

The textbook's answer reveals that the Second Partials Test fails in this case. However, the textbook also reveals that there is no relative extrema but rather absolute minima at (1, a, 0) and (b, -4, 0).

1. Why is the answer an ABSOLUTE MINIMA and not a RELATIVE EXTREMA?

2. In Terms of the absolute MINIMA, where do the letters a and b come from or can it simply be a textbook typo?

Hi Harpazo! (Smile)

When we fill in $x=1$, we get $f_x=f_y=0$, regardless of the value of $y$.
So every point with $x=1$ is a critical point - it's a vertical line.
And we have $f(x,y)=0$ at every point of that line.

Let's pick an arbitrary $y$-coordinate, say, $a$.
Then we effectively have the parametrized line $(x,y,z)=(1,a,0)$ with parameter $a$.

Similarly, the line $(x,y,z)=(b,-4,0)$ consists of critical points as well.

And yes, the second partial test fails, but we can still take a look at what the function looks like.
The function is a product of squares. That means it's either positive or zero.
It looks like:
\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[ymin=-8]
\addplot[blue, ultra thick, domain=-7:4] (1, x);
\addplot[blue, ultra thick] (x,-4);
\node at (axis cs:-2,0) {+};
\node at (axis cs:3,0) {+};
\node at (axis cs:-2,-6) {+};
\node at (axis cs:3,-6) {+};
\end{axis}
\end{tikzpicture}
In other words, those critical lines are absolute minima of the surface $z=f(x,y)$.

 

[harpazo]

Hi Harpazo! (Smile)

When we fill in $x=1$, we get $f_x=f_y=0$, regardless of the value of $y$.
So every point with $x=1$ is a critical point - it's a vertical line.
And we have $f(x,y)=0$ at every point of that line.

Let's pick an arbitrary $y$-coordinate, say, $a$.
Then we effectively have the parametrized line $(x,y,z)=(1,a,0)$ with parameter $a$.

Similarly, the line $(x,y,z)=(b,-4,0)$ consists of critical points as well.

And yes, the second partial test fails, but we can still take a look at what the function looks like.
The function is a product of squares. That means it's either positive or zero.
It looks like:
\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[ymin=-8]
\addplot[blue, ultra thick, domain=-7:4] (1, x);
\addplot[blue, ultra thick] (x,-4);
\node at (axis cs:-2,0) {+};
\node at (axis cs:3,0) {+};
\node at (axis cs:-2,-6) {+};
\node at (axis cs:3,-6) {+};
\end{axis}
\end{tikzpicture}
In other words, those critical lines are absolute minima of the surface $f(x,y)=0$.

Great reply. I need you to simply define relative extrema and absolute minima. Also, what is the difference between relative extrema and absolute extrema?

 

[I like Serena]

Great reply. I need you to simply define relative extrema and absolute minima. Also, what is the difference between relative extrema and absolute extrema?

A relative (or local) extremum is a point where all points "closeby" either have function values that are all greater (relative minimum), or all less (relative maximum).
An absolute extremum is a point such that no other point in the domain has a greater function value (absolute minimum) respectively lower (absolute maximum).

A critical point is a "candidate" for a relative extremum.
And each relative extremum is a "candidate" for an absolute extremum.

 

[harpazo]

A relative (or local) extremum is a point where all points "closeby" are either have function values that are all greater (relative minimum), or are all less (relative maximum).
An absolute extremum is a point such that no other point in the domain has a greater function value (absolute minimum) respectively lower (absolute maximum).

A critical point is a "candidate" for a relative extremum.
And each relative extremum is a "candidate" for an absolute extremum.

Good information. Thanks.