Find critical points of two equations

derivativesordinary differential equationspartial derivative

So i have this function

$$f(x, y) = (x − y)(1 − xy)$$

which i re-write as $x-x^2y-y+xy^2$

I find the partial derivatives:

$$f_x(x,y)=1-2xy+y^2$$

$$f_y(x,y)=-x^2-1+2xy$$

I want to look for points $(a,b)$ such that $f_x(a,b)=0$ and $f_y(a,b)=0$

So i will write $x$ in terms of $a$ and $y$ in terms of $b$ .

So i write:

$$1-2ab+b^2=0$$

and

$$-a^2-1+2ab=0.$$

I have dyscalculia and have no clue how to proceed from here…the values i get are weird and do not make sense…

I read somewhere that when you are not able to take the critical points of an equation , then you might want to right it in another form ? Is that possible?

Or if i'm saying nonesense, can somebody please help me and show me some simple steps on how to get to the critical points, with explanations on why do you such steps? On the answer sheet , the critical points are:

$P_1(1, 1)$ and $P_2(−1,−1)$

Thanks!!

Best Answer

From \begin{align} 1-2ab+b^2&=0\\ -a^2-1+2ab&=0 \end{align} Multiple the second by $-1$ to get

\begin{align} 1-2ab+b^2&=0\\ a^2+1-2ab&=0 \end{align} Now notice that $1 - 2ab$ appears in both, so write \begin{align} 1-2ab&= -b^2\\ 1-2ab&=-a^2 \end{align}

from which you can conclude that $a^2 = b^2$, which gives you two cases:

  1. $a = b$
  2. $a = -b$.

For the first, your first equation becomes \begin{align} 1-2ab+b^2&=0\\ 1-2a^2+a^2&=0\\ 1-a^2&=0\\ a^2&=1\\ a&=\pm 1 \end{align} and hence leads to the solutions $(1,1)$ and $(-1, -1)$.

Now YOU try to do the same thing for the second case, substituting $-a$ for $b$ in the equations to see what you get.

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