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 Calculus Calculus Math Forum

 August 3rd, 2017, 05:20 AM #1 Member   Joined: May 2017 From: Slovenia Posts: 89 Thanks: 0 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  August 3rd, 2017, 12:09 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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 $\displaystyle a^2+ 5a- 8= 0$. Thanks from greg1313 and sarajoveska Tags extrema, local Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post sarajoveska Calculus 1 July 13th, 2017 01:41 PM ZMD Linear Algebra 1 April 29th, 2017 04:57 PM Mrto Calculus 3 April 23rd, 2016 03:26 PM CPAspire Pre-Calculus 2 March 27th, 2015 07:52 AM crnogorac Calculus 1 December 24th, 2013 04:03 AM

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