August 3rd, 2017, 05:20 AM  #1 
Member Joined: May 2017 From: Slovenia Posts: 52 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: 2,576 Thanks: 667 
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 $\displaystyle a^2+ 5a 8= 0$. 

Tags 
extrema, local 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Local extrema  sarajoveska  Calculus  1  July 13th, 2017 01:41 PM 
Local extrema  ZMD  Linear Algebra  1  April 29th, 2017 04:57 PM 
Local extrema  Mrto  Calculus  3  April 23rd, 2016 03:26 PM 
Proving local extrema using variables.  CPAspire  PreCalculus  2  March 27th, 2015 07:52 AM 
Local Extrema  crnogorac  Calculus  1  December 24th, 2013 04:03 AM 