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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
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Re: Stationary points ..

Hello, wulfgarpro!

Quote:
 Find the stationary points of:[color=beige] .[/color] I found the partial derivatives: [color=beige]. . [/color]

[color=beige]. . . . [/color]

[color=beige]. . [/color]I'll let you classify them . . . May 28th, 2010, 01:31 PM   #3
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Re: Stationary points ..

Hm.

Quote:
 Originally Posted by soroban [color=beige]. . . . [/color]
Hi Soroban,

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: 20,747 Thanks: 2133 2y - x� = 0 x - 4y� = 0 Hence 2y = x� = (4y�)� = 16y^4. This is clearly true if (x, y) = (0, 0). If y is not zero, dividing by 2y gives 1 = 8y�, 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: 20,747 Thanks: 2133 (1/2)� = 1/2� = 1/8. Alternatively, y� = 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. Tags points, stationary 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 iulia.d Calculus 0 March 20th, 2014 12:48 AM ARTjoMS Calculus 2 January 9th, 2013 11:59 AM sahil112 Calculus 5 November 21st, 2012 11:51 AM Voltman Algebra 2 June 15th, 2011 06:22 PM jakeward123 Calculus 8 May 4th, 2011 01:51 AM

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