Stationary Point Question?

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September 20th 2010, 09:45 PM
Paymemoney

Stationary Point Question?

Hi
Can someone tell me if i have done the following question correctly because according to the book's answers this is incorrect.

For the surface given $z=4x^3-12x+y^2+8y+8$
Find the stationary point or points and determine the nature of any such point.

$\frac{dz}{dx} = 0$
$\frac{dz}{dy} = 0$

$0=12x^2-12$

$x=1$

$0=2y+8$

$y=-4$

$z=-16$

$\frac{\partial^2 z}{\partial x^2} = 12x$

$\frac{\partial^2 z}{\partial y^2} = 2$

$\frac{\partial^2 z}{\partial x \partial y } = 0$

because there is still $12x$ i differential again

$\frac{\partial^3 z}{\partial x^3} = 12$

$\frac{\partial^3 z}{\partial y^3} = 0$

$\frac{\partial^3 z}{\partial x \partial y } = 0$

$g= 12*0-0$

$g=0$

therefore i get saddle point according to the statement

Points (1,-4,-16) $\frac{\partial^2 z}{\partial x^2}$ > 0 & g > 0

answers is (1,-2,-8) is a local minmum; (-1,-2, 8) is a saddle point

how do i find the minmum point?

P.S
September 20th 2010, 10:04 PM
mr fantastic

Quote:

Originally Posted by Paymemoney

Hi
Can someone tell me if i have done the following question correctly because according to the book's answers this is incorrect.

For the surface given $z=4x^3-12x+y^2+8y+8$
Find the stationary point or points and determine the nature of any such point.

$\frac{dz}{dx} = 0$
$\frac{dz}{dy} = 0$

$0=12x^2-12$

$x=1$ Mr F says: You have missed the other solution to this equation ........ (I wish I had a dollar for every time this careless mistake was made).

$0=2y+8$

$y=-4$ Mr F says: Correct.

$z=-16$

$\frac{\partial^2 z}{\partial x^2} = 12x$

$\frac{\partial^2 z}{\partial y^2} = 2$

$\frac{\partial^2 z}{\partial x \partial y } = 0$

because there is still $12x$ i differential again

$\frac{\partial^3 z}{\partial x^3} = 12$

$\frac{\partial^3 z}{\partial y^3} = 0$

$\frac{\partial^3 z}{\partial x \partial y } = 0$

$g= 12*0-0$

$g=0$

therefore i get saddle point according to the statement

Points (1,-4,-16) $\frac{\partial^2 z}{\partial x^2}$ > 0 & g > 0

answers is (1,-2,-8) is a local minmum; (-1,-2, 8) is a saddle point

how do i find the minmum point?

P.S

The test for the nature of the critical point is given here: http://noether.uoregon.edu/~dps/242/Notes/notes21.pdf
September 20th 2010, 10:18 PM
Paymemoney

well if one of the points in -4 then the book's answer must be wrong.