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May 28th, 2010, 02:30 AM  #1 
Senior Member Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0  Stationary points ..
Hi, Looking at some stationary point work. I'm stumped. z = f(x, y) = 6xy  x^(3)  8y^(3) I have found the partial derivative for f(x, y) in respect to firstly x, and then y as: f_x = 6y  3x^(2) f_y = 6x  24y^(2) By the theory of stationary points, these derivatives need to be solved to equate to 0. Therefore, I need to find solutions for f_x = 0 and f_y = 0. I have simplified the equations to be: 2y  x^(2) = 0 x  4y^(2) = 0 Now, my professor gave me some work from here, but I am still trying to understand it and would appreciate any assistance: Firstly, as we are trying to solve an algebraic equation of two variables, we can say: 2y = x^(2) = 4y^(2) = 16y^(4) So, 2y = 16y^(4), therefore, y = 8y(^4) After this, he loses me: 8y^(4)  y = 0; therefore, y = 0; OR; 8y^(3)  1 = 0, i.e. y^(3) = 1/8; therefore, y = 1/2 Thus, there are 2 solutions for y: y = 0 and y = 1/2. How does he come to 8y^(4)  y = 0 AND 8y^(3)  1 = 0 ??  Any help would be greatly appreciated! wulfgarpro. 
May 28th, 2010, 06:13 AM  #2  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Re: Stationary points .. Hello, wulfgarpro! Quote:
[color=beige]. . . . [/color] [color=beige]. . [/color]I'll let you classify them . . .  
May 28th, 2010, 01:31 PM  #3  
Senior Member Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0  Re: Stationary points ..
Hm. Quote:
Thank you for your reply! Could you explain to me exactly what you are doing in the above. wulfgarpro.  
May 29th, 2010, 02:13 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,715 Thanks: 1806 
2y  x² = 0 x  4y² = 0 Hence 2y = x² = (4y²)² = 16y^4. This is clearly true if (x, y) = (0, 0). If y is not zero, dividing by 2y gives 1 = 8y³, which leads to (x, y) = (1, 1/2). 
May 29th, 2010, 06:35 PM  #5 
Senior Member Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0  Re: Stationary points ..
Hi skipjack, How does this work: y^(3) = 1/8 y = 1/2 ? wulfgarpro. 
May 30th, 2010, 02:14 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,715 Thanks: 1806 
(1/2)³ = 1/2³ = 1/8. Alternatively, y³ = 1/8 implies y = (1/^(1/3), which turns out to be 1/2 when evaluated using a calculator. However the previous method shows that the answer is exactly 1/2. 

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