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May 28th, 2010, 03:30 AM   #1
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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.
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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]

I found the partial derivatives:

[color=beige]. . [/color]





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






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

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May 28th, 2010, 02:31 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.
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May 29th, 2010, 03:13 PM   #4
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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).
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May 29th, 2010, 07:35 PM   #5
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Re: Stationary points ..

Hi skipjack,

How does this work:

y^(3) = 1/8
y = 1/2

?

wulfgarpro.
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May 30th, 2010, 03:14 AM   #6
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(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.
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