Thread: Another local extreme problem

I am facing another problem that I can't compute the extremes forDoh)

f(xy)=e^x(x²-y²)

so far I have:

fx= 2xe^x + e^x(x²-y²)
fy= -2ye^x
fxx=2e^x+2xe^x + e^x(x²-y²) + 2xe^x
fyy= -2e^x
fxy= -2ye<sup>x

How do I find the extremes? I set fx and fy to 0, but I can't solve for x and y.

Thanks for any help.</sup>

Originally Posted by MajMinor

I am facing another problem that I can't compute the extremes forDoh)

f(xy)=e^x(x²-y²)

so far I have:

fx= 2xe^x + e^x(x²-y²)
fy= -2ye^x
fxx=2e^x+2xe^x + e^x(x²-y²) + 2xe^x
fyy= -2e^x
fxy= -2ye<sup>x

How do I find the extremes? I set fx and fy to 0, but I can't solve for x and y.

Thanks for any help.</sup>

If you set the partial with respect to x equal to 0 you get




Since the exponential is never zero his gives



If we plug this into the 2nd equation we get





Again the exponential can never be zero so we get




So we get the two solutions



Can you finish from here?

Yes. Thank you very much! It's the little things that always trip me up.