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 Post subject: Solving for X,Y (Part of a critical Points problem)Posted: Thu, 10 May 2012 23:09:18 UTC
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Joined: Thu, 10 May 2012 22:51:47 UTC
Posts: 2
First of this is somewhat embarassing as I really shouldn't be strugling with this, however I'm having a hard time getting back into this after a break of about a year.

I'm trying to obtain the critical points of the function bellow and then determin their nature. I'm having no trouble with the differentiation however solving the first derivatives for x and y is driving me nuts. Once I have these values I know how to solve the rest of the problem.

A surface has the following equation:

z=x^3-3xy+8y^3+20

So:

dz/dx= 3x^2-3y = 0

dz/dy= -3x+24y^2 = 0

thus:

3x^2=3y so x^2=y

and

24y^2=3x so 8y^2=x

CP's:

x = (A)??? & (B)???
y = (A)??? & (B)???

Z= (A)x^3-3(A)x(A)y+8(A)y^3+20
&
Z= (B)x^3-3(B)x(B)y+8(B)y^3+20

Any help is greatly appreciated.

Craig

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 Posted: Fri, 11 May 2012 00:20:18 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 14003
Location: Austin, TX
Phoenix86 wrote:
First of this is somewhat embarassing as I really shouldn't be strugling with this, however I'm having a hard time getting back into this after a break of about a year.

I'm trying to obtain the critical points of the function bellow and then determin their nature. I'm having no trouble with the differentiation however solving the first derivatives for x and y is driving me nuts. Once I have these values I know how to solve the rest of the problem.

A surface has the following equation:

z=x^3-3xy+8y^3+20

So:

dz/dx= 3x^2-3y = 0

dz/dy= -3x+24y^2 = 0

thus:

3x^2=3y so x^2=y

and

24y^2=3x so 8y^2=x

CP's:

x = (A)??? & (B)???
y = (A)??? & (B)???

Z= (A)x^3-3(A)x(A)y+8(A)y^3+20
&
Z= (B)x^3-3(B)x(B)y+8(B)y^3+20

Any help is greatly appreciated.

Craig

You have a critical point when the tangent plane is "vertical", so first you find the equation for your tangent plane:

Write then the tangent plane is

now and similarly, , and the plane is vertical when both are equal to zero.

So all you need to do is decide when

Using the second equation and squaring we get:

and substituting into the first you get:

, so we get which works in the original set of equations, otherwise so , this gives the other option as:

so and from the first equation, and checking with the second equation, we see that only the positive possibility works.

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 Posted: Fri, 11 May 2012 13:38:06 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
You have a critical point when the tangent plane is "vertical".

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 Posted: Fri, 11 May 2012 13:44:34 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 14003
Location: Austin, TX
outermeasure wrote:
You have a critical point when the tangent plane is "vertical".

Probably, I had a mnemonic which is technically incorrect, but always lets me remember how it goes which involved a relation to vertical lines in the case.

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 Posted: Sat, 12 May 2012 22:37:02 UTC
 S.O.S. Newbie

Joined: Thu, 10 May 2012 22:51:47 UTC
Posts: 2
Thanks for that, was starting to get kind of worried when I couldnt figure that out... seems kinda obvious now

Craig

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