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I have this function:

$$ f(x,y) = x^2 - 2xy+ 4y^3$$

I calculated the gradient without problems:

$$\nabla f(x,y) = \left(2x-2y , -2x + 12y^3\right)^T$$

This is where kind of got stuck, I know from looking at the gradient that for $(0,0)$ the gradient is $0$, so we have a critical point there but, other than that I'm lost.

I know from $ 2x-2y$ that for for the first part of the gradient to be zero $x$ has to be equal to $y$.

I dont know how to find out the other critical points.

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I have this function: $$ f(x,y) = x^2 - 2xy+ 4y^3$$ I calculated the gradient without problems: $$\nabla f(x,y) = \left(2x-2y , -2x \color{red}{-} 12y^\color{red}{3}\right)^T$$

Careful, you have a small sign and exponent mistake: $\left(2x-2y , -2x \color{green}{+} 12y^\color{green}{2}\right)^T$.

I dont know how to find out the other critical points.

But you haven't found any critical points yet, as you only used one equation so far. Keep in mind that requiring that the gradient is equal to $(0,0)$ is equivalent to a system of two equations: $$\left\{ \begin{array}{rcl} 2x-2y &=& 0 \\ -2x + 12y^2 &=& 0 \end{array}\right. \iff \left\{ \begin{array}{rcl} x &=& y \\ -2x + 12y^2 &=& 0 \end{array}\right.$$ Now plug $x=y$ into the second equation and solve: $$-2x + 12x^2 = 0 \iff -2 x\left( 1 -6x \right) = 0 \iff x = \ldots$$ Since $y=x$, each solution $x$ will correspond to a critical point $(x,x)$. Can you finish?

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  • $\begingroup$ thanks for the correction, I think i solved it now with $x = \frac{1}{6}$ and $x=0$ $\endgroup$ – zython Aug 17 '17 at 14:00
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    $\begingroup$ @zython Correct, but note that the critical points (since $y=x$) are then $(0,0)$ and $(\tfrac{1}{6},\tfrac{1}{6})$. $\endgroup$ – StackTD Aug 17 '17 at 14:01
  • $\begingroup$ this way of solving this feels a little bit strange is there a better way of doing this ? $\endgroup$ – zython Aug 17 '17 at 14:02
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    $\begingroup$ @zython What exactly do you find strange about it? You need to find the points for which the gradient is the zero vector, so this will come down to solving a system of equations. You can solve the system in any way you like, but using substitution is an obvious choice here. $\endgroup$ – StackTD Aug 17 '17 at 14:04
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    $\begingroup$ Do you mean the part of solving the system? I'm afraid that heavily depends on the system. There are more algorithmic methods for linear systems, but the system won't be linear in general (as in this example). As is often the case, experience & exercise will increase insight and speed up calculations. $\endgroup$ – StackTD Aug 17 '17 at 14:11
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You are right that you want the gradient to be the zero vector:

Both $2x-2y=0$ and $-2x-12y^3=0$

Starting with $2x-2y=0$ gives $x=y$.

Assuming that $x=y$, turns $-2x-2y^3=0$ into $-2y-12y^3=0$.

We can solve this by factorising: $-2y(1+6y^2)=0$.

Once you solve this for $y$, your corresponding $x$-value comes from knowing that $x=y$.

Note: There is only one critical point if you are working over the reals. There are three critical points if $x$ and $y$ are allowed to be complex.

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Yes, in order to obtain the critical points of $f(x,y) = x^2 - 2xy+ 4y^3$ you have to solve $$\nabla f(x,y) =\left(f_x(x,y) ,f_y(x,y)\right)= \left(2x-2y , -2x + 12y^2\right)=(0,0).$$ Note the above gradient is different from yours!

From $2x-2y=0$ we have that $x=y$. Then plugging this one into $-2x + 12y^2=0$ we get $$-2x + 12x^2=2x(-1+6x)=0\implies x=0 \;\mbox{or}\;x=1/6.$$ Therefore the critical points are $(0,0)$ and $(1/6,1/6)$.

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  • $\begingroup$ I have calculated solutions $ x = \frac{1}{6} $ and $x=0$, is that correct ? $\endgroup$ – zython Aug 17 '17 at 13:58
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    $\begingroup$ @zython Yes, therefore the critical points are $(0,0)$ and $(1/6,1/6)$. $\endgroup$ – Robert Z Aug 17 '17 at 13:59

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