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August 3rd, 2017, 06:20 AM   #1
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Local extrema

I need to find the parameter a so that the function f(x,y)=e^(ax)(ax+y^2+4y) has at the point T (1, -2) local extrema. I'm really confused with this one, any help will be appreciated
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August 3rd, 2017, 01:09 PM   #2
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Set the partial derivatives to 0:
With $\displaystyle f(x)= e^{ax}(ax+y^2+4y)$
$\displaystyle f_x= ae^{ax}(ax+ y^2+ 4y)+ e^{ax}(a)= e^{ax}(a^2x^2+ a+ ay^2+ 4y)= 0$
$\displaystyle f_y= e^{ax}(2y+ 4)= 0$

Since $\displaystyle e^{ax}$ is never 0, we must have $\displaystyle a^2x^2+ a+ ay^2+ 4y= 0$ and $\displaystyle 2y+ 4= 0$.

At x= 1, y= -2, the first equation becomes $\displaystyle a^2+ a+ 4a- 8= a^2+ 5a- 8= 0$ while the second equation becomes just $\displaystyle -4+ 4= 0$. You need to solve the quadratic equation $\displaystyle a^2+ 5a- 8= 0$.
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