find the stationary point for $f(x,y) =x^4 +64y^4-2(x+8y)^2$
i got my fx =3x^3 -4x-32y
and fy= 256y^3-32x-256y

I don't know how to get the three points....any one can highlight to me??

2. Hello

Hint : $\frac{32}{4}=\frac{256}{32}$ It should help you getting rid off $x$ and $y$ and you'll be able to solve for $x^3$ and $y^3$

3. Hello,

fx would rather be $4x^3-\dots$

Stationary points are points (x,y) where $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$

So solve for x and y in

$\bigg\{ \begin{array}{ccc} 4x^3-4x-32y & = & 0 \\ 256y^3-32x-256y & = & 0 \end{array}$

4. stuck

$y^3-32x/256-y=0$
$x^3-x-32y/4=0$ how to proceed?

5. I think that the better way is, as Flyingsquirrel mentioned, to see that 256/32=32/4=8

So if you multiply the first equation by 8, you will be able to get rid of x and y, and keep y^3, while substracting term by term the two equations

6. stuck again

2y=x i got it after some manipulation........what next????

7. What about substituting $x=2y$ in one of the equation and solving for $y$ ?

8. stuck again again

sub x=2y into the f(x,y)
i got 80y^4 - 200y^2

9. "Substituting in one of the equations" meant substituting either in $f_x=0$ or in $f_y=0$ but not in $f(x,y)$, which is not an equation.

10. i got y=+-(5/4)^(1/2)

11. No, these are not the solutions for $y$.

$4x^3-4x-32y=0 \Rightarrow 4\times8y^3-8y-32y=0 \Rightarrow 16y(2y^2-3)=0 \Rightarrow \left\{ \begin{array}{l}y=\ldots\\ \mbox{or} \\y=\ldots\\ \mbox{or} \\y=\ldots\end{array}\right.$

12. solve

ehhh....really??
i better look at it....

13. equation correct?

$8y^3 - x- 8y$

$x^3-x-8y$

two equation correct?

14. Yes

15. solve

solve....=)

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