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 May 28th, 2010, 03:30 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, 07:13 AM   #2
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Re: Stationary points ..

Hello, wulfgarpro!

Quote:
 Find the stationary points of:[color=beige] .[/color]$f(x, y) \:=\: 6xy \,-\,x^3 \,-\, 8y^3$ I found the partial derivatives: [color=beige]. . [/color]$\begin{array}{ccccc}f_x=&\,=\,&6y\,-\,3x^2=&\,=\, &0 \\ \\ f_y=&\,=\, &6x\,-\,24y^2=&\,=\, &0 \end{array}=$

$\text{We have: }\;\begin{array}{cccccc}2y=&x^2=&\;[1] \\ x=&4y^2=&\;[2] \end{array}=$

$\text{Substitute [2] into [1]: }\;2y \;=\;\left(4y^2\right)^2 \;\;\;\Rightarrow\;\;\;2y \:=\:16y^4$

[color=beige]. . . . [/color]$16y^4\,-\,y \:=\:0 \;\;\;\Rightarrow\;\;\;2y(8y^3\,-\,1) \:=\:0 \;\;\;\Rightarrow\;\;\;y \:=\:0,\:\frac{1}{2}$

$\text{Substitute into [2]: }\;x \;=\;0,\:1$

$\text{Stationary points: }\;(0,\:0,\:0)\,\text{ and }\,\left(1,\:\frac{1}{2},\:1\right)$

[color=beige]. . [/color]I'll let you classify them . . .

May 28th, 2010, 02:31 PM   #3
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From: Melbourne

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Re: Stationary points ..

Hm.

Quote:
 Originally Posted by soroban [color=beige]. . . . [/color]$16y^4\,-\,y \:=\:0 \;\;\;\Rightarrow\;\;\;2y(8y^3\,-\,1) \:=\:0 \;\;\;\Rightarrow\;\;\;y \:=\:0,\:\frac{1}{2}$
Hi Soroban,

Thank you for your reply! Could you explain to me exactly what you are doing in the above.

wulfgarpro.

 May 29th, 2010, 03:13 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,270 Thanks: 1958 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, 07:35 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, 03:14 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,270 Thanks: 1958 (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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