# Finding critical numbers of a product involving an exponential function

• March 3rd 2008, 02:16 PM
emttim84
Finding critical numbers of a product involving an exponential function
Ok, so the original function is $f(x,y)=(x^2+4y^2)e^{1-x^2-y^2}$ and the problem is to find the derivative with respect to x, respect to y, respect to xx, respect to yy, and respect to xy/yx and find the critical numbers as well as any relative extrema or saddle point if applicable.

I found that the derivative with respect to x is $f'x(x,y)=2xe^{1-x^2-y^2}(1-x^2-4y^2)$, the derivative with respect to y is $f'y(x,y)=2ye^{1-x^2-y^2}(4-x^2-4y^2)$, the derivative with respect to xx is $f'xx(x,y)=2e^{1-x^2-y^2}(1-5x^2+2x^4+8x^2y^2-4y^2)$, the derivative with respect to yy is $f'yy(x,y)=2e^{1-x^2-y^2}(4-x^2-20y^2+2x^2y^2+8y^4)$ and finally the derivative with respect to xy is $f'xy(x,y)=-4xye^{1-x^2-y^2}(5-x^2-4y^2)$.

Now, I tried setting $2xe^{1-x^2-y^2}$ from $f'x$ equal to $2ye^{1-x^2-y^2}$ from $f'y$ but that didn't work. I tried doing substitution for $4-x^2-4y^2=0$ and $1-x^2-4y^2=0$ but that ends up being 1=4 which obviously isn't true nor is it a CN...so how would I find the CN's for this problem? I'm sure I'm overlooking something simple, however, I can't find out what that simple thing is.
• March 3rd 2008, 03:22 PM
mr fantastic
Quote:

Originally Posted by emttim84
Ok, so the original function is $f(x,y)=(x^2+4y^2)e^{1-x^2-y^2}$ and the problem is to find the derivative with respect to x, respect to y, respect to xx, respect to yy, and respect to xy/yx and find the critical numbers as well as any relative extrema or saddle point if applicable.

I found that the derivative with respect to x is $f'x(x,y)=2xe^{1-x^2-y^2}(1-x^2-4y^2)$, the derivative with respect to y is $f'y(x,y)=2ye^{1-x^2-y^2}(4-x^2-4y^2)$, the derivative with respect to xx is $f'xx(x,y)=2e^{1-x^2-y^2}(1-5x^2+2x^4+8x^2y^2-4y^2)$, the derivative with respect to yy is $f'yy(x,y)=2e^{1-x^2-y^2}(4-x^2-20y^2+2x^2y^2+8y^4)$ and finally the derivative with respect to xy is $f'xy(x,y)=-4xye^{1-x^2-y^2}(5-x^2-4y^2)$.

Now, I tried setting $2xe^{1-x^2-y^2}$ from $f'x$ equal to $2ye^{1-x^2-y^2}$ from $f'y$ but that didn't work. I tried doing substitution for $4-x^2-4y^2=0$ and $1-x^2-4y^2=0$ but that ends up being 1=4 which obviously isn't true nor is it a CN...so how would I find the CN's for this problem? I'm sure I'm overlooking something simple, however, I can't find out what that simple thing is.

When you put each of your (correct (Clapping) ) partial derivatives equal to zero you get:

x = 0 or $1 - x^2 - 4y^2 = 0$

y = 0 or $4 - x^2 - 4y^2 = 0$

Note that the factor $e^{1 - x^2 - y^2} \neq 0$ for real values of x and y.

You're quite right about $1 - x^2 - 4y^2 = 0$ and $4 - x^2 - 4y^2 = 0$ being simultaneously inconsistent.

But it looks like you overlooked the simple solution (0, 0) ........
• March 3rd 2008, 08:03 PM
emttim84
Quote:

Originally Posted by mr fantastic
When you put each of your (correct (Clapping) ) partial derivatives equal to zero you get:

x = 0 or $1 - x^2 - 4y^2 = 0$

y = 0 or $4 - x^2 - 4y^2 = 0$

Note that the factor $e^{1 - x^2 - y^2} \neq 0$ for real values of x and y.

You're quite right about $1 - x^2 - 4y^2 = 0$ and $4 - x^2 - 4y^2 = 0$ being simultaneously inconsistent.

But it looks like you overlooked the simple solution (0, 0) ........

Hmm, fair enough, but the solutions manual states there are the critical numbers 0,0....0,1...0,-1...1,0...-1,0 so I'm mainly unsure of where they get the plus or minus 1 for the X and then the Y values from.
• March 3rd 2008, 08:19 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
When you put each of your (correct (Clapping) ) partial derivatives equal to zero you get:

x = 0 or $1 - x^2 - 4y^2 = 0$

y = 0 or $4 - x^2 - 4y^2 = 0$

Note that the factor $e^{1 - x^2 - y^2} \neq 0$ for real values of x and y.

You're quite right about $1 - x^2 - 4y^2 = 0$ and $4 - x^2 - 4y^2 = 0$ being simultaneously inconsistent.

But it looks like you overlooked the simple solution (0, 0) ........

The other possibilities for the partial derivatives being equal to zero are:

Case 1. x = 0 and $4 - x^2 - 4y^2 = 0$.

Case 2. y = 0 and $1 - x^2 - 4y^2 = 0$.

Case 1: Substitute x = 0 into $4 - x^2 - 4y^2 = 0$ and you have $4 - 4y^2 = 0$ ......

So (0, 1) and (0, -1) are critical points.

Case 2: Substitute y = 0 into $1 - x^2 - 4y^2 = 0$ and you have .......
• March 3rd 2008, 08:28 PM
emttim84
Quote:

Originally Posted by mr fantastic
The other possibilities for the partial derivatives being equal to zero are:

Case 1. x = 0 and $4 - x^2 - 4y^2 = 0$.

Case 2. y = 0 and $1 - x^2 - 4y^2 = 0$.

Case 1: Substitute x = 0 into $4 - x^2 - 4y^2 = 0$ and you have $4 - 4y^2 = 0$ ......

So (0, 1) and (0, -1) are critical points.

Case 2: Substitute y = 0 into $1 - x^2 - 4y^2 = 0$ and you have .......

Ah-ha....I see. So if I get a problem like this where a little digging must be done to find the CN's, assume x=0 for the derivative of a function with respect to x and plug x=0 into the derivative to find some of the CN's and assume y=0 for the derivative of a function with respect to y and plug in y=0 into the derivative to find the rest of the CNs? Thanks!
• March 3rd 2008, 08:42 PM
mr fantastic
Quote:

Originally Posted by emttim84
Ah-ha....I see. So if I get a problem like this where a little digging must be done to find the CN's, assume x=0 for the derivative of a function with respect to x and plug x=0 into the derivative to find some of the CN's and assume y=0 for the derivative of a function with respect to y and plug in y=0 into the derivative to find the rest of the CNs? Thanks!

$f_x = 0$ when x = 0, $f_y = 0$ when y = 0. There's no assumption .....

When $f_x = 0$ it's a matter of looking for when $f_y = 0$ .......

Ditto when $f_y = 0$ .....