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March 9th, 2014, 09:08 AM   #1
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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
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 March 9th, 2014, 09:51 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: 2x-1/(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, 02:06 PM   #3
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Joined: Feb 2014

Posts: 37
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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...
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 March 9th, 2014, 03:00 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, 06:19 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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