How to Find a Region Where V'(x,y) < 0?

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SUMMARY

The discussion focuses on finding a region where the derivative of the function V(x,y) = 0.45x² + xy + 0.56y² is negative, specifically V'(x,y) = -8.4840x² - 18.7412x - 10.3496 - 0.011xy - 0.011y². The key method suggested for this analysis is completing the square to rewrite the expression in a more manageable form. This approach will help identify the neighborhood where V'(x,y) < 0, ensuring the function remains positively invariant.

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Somefantastik
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Hi,

I've got a function that I'm trying to show is positively invarient.

[tex]Let \ V(x,y) = 0.45x^{2}+xy+0.56y^{2} > 0 \ for \ all \ x,y \ in \ R^{2}[/tex]

[tex]V'(x,y) = ... = -8.4840x^{2} - 18.7412x - 10.3496 - 0.011xy - 0.011y^{2};[/tex]

How can I find a neighborhood/region that makes V'(x,y)< 0?
 
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Hi Somefantastik! :smile:

Complete the square (twice) … write it in the form -A(x - B)2 - C(x - D)2 +E :wink:
 

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