Find the stationary points of this function f(x)=x^3 - 6xy + 3y^2

Can someone please help me with calculus problem? (detailed please)

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- Dec 12th 2010, 03:31 AMAhplymFind the stationary points.
Find the stationary points of this function f(x)=x^3 - 6xy + 3y^2

Can someone please help me with calculus problem? (detailed please) - Dec 12th 2010, 04:07 AMProve It
Your function is $\displaystyle \displaystyle f(x,y) = x^3 - 6xy + 3y^2$.

The gradient vector is $\displaystyle \displaystyle \nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$ and the critical points occur when $\displaystyle \displaystyle \nabla f = \mathbf{0}$.

$\displaystyle \displaystyle \nabla f = \left(3x^2 - 6y, -6x + 6y\right) = \mathbf{0}$.

So $\displaystyle \displaystyle 3x^2 - 6y = 0$ and $\displaystyle \displaystyle -6x + 6y = 0$.

From the second equation, $\displaystyle \displaystyle x = y $.

Substituting into the first gives $\displaystyle \displaystyle 3x^2 - 6x = 0$

$\displaystyle \displaystyle 3x(x - 2) = 0$

$\displaystyle \displaystyle x = 0, x = 2$.

Therefore, the critical points are $\displaystyle \displaystyle (x, y) = (0, 0)$ and $\displaystyle \displaystyle (x, y) = (2, 2)$. - Dec 12th 2010, 04:19 AMAhplym
Thank you so much.