MHB Why does the following have no maximums or minimums

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I have the following function:

p(x,y) = (2-x2) exp(-y)

I have found the partial derivatives:
dp/dx = 2x*exp(-y)

dp/dy = -2*exp(-y) + x2exp(-y)

By solving these i found the points x = 0, and \pm 20.5.However, the answer i have been given is that there is no maximums and minimums.

Can someone please clarify where i am going wrong.

Thanks in advance (Happy)
 
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Check your result for $$\pd{p}{x}$$...and then equating the partials you should find you have a quadratic in $x$ with a negative discriminant.
 
brunette15 said:
I have the following function:

p(x,y) = (2-x2) exp(-y)

I have found the partial derivatives:
dp/dx = 2x*exp(-y)

dp/dy = -2*exp(-y) + x2exp(-y)

By solving these i found the points x = 0, and \pm 20.5.However, the answer i have been given is that there is no maximums and minimums.

Can someone please clarify where i am going wrong.

Thanks in advance (Happy)
Because $e^{- y}$ never vanishes you can divide by it and the conditions fom maximum or minimum become...

$\displaystyle 2\ x = x^{2}-2 = 0\ (1)$

... and there is no value of x satisfiyng (1)...

Kind regards

$\chi$ $\sigma$
 
brunette15 said:
I have the following function:

p(x,y) = (2-x2) exp(-y)

I have found the partial derivatives:
dp/dx = 2x*exp(-y)

dp/dy = -2*exp(-y) + x2exp(-y)

By solving these i found the points x = 0, and \pm 20.5.However, the answer i have been given is that there is no maximums and minimums.

Can someone please clarify where i am going wrong.

Thanks in advance (Happy)

You have a small error. $\displaystyle \begin{align*} \frac{\partial p}{\partial x } = -2x\,\mathrm{e}^{-y} \end{align*}$, not $\displaystyle \begin{align*} 2x\,\mathrm{e}^{-y} \end{align*}$...
 

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