Why does the following have no maximums or minimums

[brunette15]

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 2^0.5.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)

 

[MarkFL]

Check your result for \(\displaystyle \pd{p}{x}\)...and then equating the partials you should find you have a quadratic in $x$ with a negative discriminant.

 

[chisigma]

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 2^0.5.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

$\chi$ $\sigma$

 

[Prove It]

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 2^0.5.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 $\displaystyle \begin{align*} 2x\,\mathrm{e}^{-y} \end{align*}$...