March 9th, 2014, 10:08 AM  #1 
Member Joined: Feb 2014 Posts: 37 Thanks: 0  maximum and minimum of a function
Dear Mentor & Guru, Help needed solve equation that attached to determine whether they are relative maximum, relative minimum or saddle points. Million Thanks 
March 9th, 2014, 10:51 AM  #2 
Newbie Joined: Mar 2014 Posts: 9 Thanks: 0  Re: maximum and minimum of a function
I'll do the first one... and you can do the second on your own. First take partial derivative with respect to x, and set it to equal to 0: 2x1/(4y) = 0 x=1/(8y) ... (first equation) Do the same thing with respect to y: 2 + x/(4y^2) = 0 Substitute x from the first equation: 2 + 1/(32y^3) = 0 y^3 = 64 y=4, and x=1/32... This is your critical point coordinate. To see what kind of points there, look here: http://en.wikipedia.org/wiki/Second_par ... ative_test 
March 9th, 2014, 03:06 PM  #3 
Member Joined: Feb 2014 Posts: 37 Thanks: 0  Re: maximum and minimum of a function
Dear Mentor & Guru, Million thanks for your guide and help to solve the question 1. I done the question 2 but not sure correct or not help to correct it if wrong and I don't know and understand how to determine whether they are relative maximum, relative minimum or saddle points. Please Help... 
March 9th, 2014, 04:00 PM  #4 
Newbie Joined: Mar 2014 Posts: 9 Thanks: 0  Re: maximum and minimum of a function
Partial derivative with respect to x: 6x^2 + 14x Set it equal to 0: 6x^2+14x=0 2x(3x + 7)=0 2x=0 > x=0 3x+7 > x=7/3 Partial derivative with respect to y: 8y Set it equal to 0: 8y=0 > y=0 Coordinates of critical points: (0,0) and (7/3,0) Take the partial derivative with respect to x, and then take its partial derivative with respect to x again, and let it be A. > A=12x+14 Take the partial derivative with respect to x, and then take its partial derivative with respect to y, and let it be B. > B = 0 Take the partial derivative with respect to y, and then take its partial derivative with respect to y again, and let it be C. > C = 8 Nature of stationary points (plug in coordinates inside A, B, and C): if AC  B^2 > 0 and A<0 , then local maximum if AC  B^2 > 0 and A>0, then local minimum if AC  B^2 < 0, then saddle point if AC  B^2 = 0, undetermined Point (0,0): AC  B^2 = (12(0)+14)(  0 < 0 This means it's a saddle point. Point (7/3,0) AC  B^2 = (12(7/3)+14)(  0 > 0, and A = 12(7/3)+14 < 0 This means it's a local maximum. 
March 9th, 2014, 07:19 PM  #5 
Member Joined: Feb 2014 Posts: 37 Thanks: 0  Re: maximum and minimum of a function
Dear Guru & Mentor, Very appreciate that provide the full detail solution, Million Thanks. I'm more understand now to solve this equation. Thanks again and apologies if caused a lot trouble for you all. 

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