Suppose you have the function, $$\space f(x,y) =5xy(9-x-y)$$ and you have to find the critical points of $\space f(x,y)$. To start, I found the $f_x$ and $f_y$ and set each to $0$. Then, I tried to solve the system of equations:

$$f_{x} = -5y^2-10xy+45y = 0$$ $$f_y = -5x^2-10xy+45x = 0$$

At this point, is there a simple method to solve this system? I am thinking I should multiply $\space f_y$ by $-1$, yielding: $$-5y^2-10xy+45y = 0$$ $$5x^2+10xy-45x = 0$$ and then adding the equations together, but I don't think that is useful. Additionally, I think I can solve the top equation for $x$ and then plug the resulting equation into $f_y$, but that is tedious.


  • $\begingroup$ $$(x=0\land y=9)\lor (x=3\land y=3)\lor (x=9\land y=0)\lor (y=0\land x=0)$$ $\endgroup$ – Moo Oct 17 '18 at 3:45
  • $\begingroup$ @Moo Oh! How did you do that? $\endgroup$ – Art Oct 17 '18 at 3:47
  • 2
    $\begingroup$ Solve the first equation for $x$ and substitute into the second. That gives you a quadratic equation. $\endgroup$ – Moo Oct 17 '18 at 3:54
  • $\begingroup$ @Moo Ok. Thanks. $\endgroup$ – Art Oct 17 '18 at 4:10

I think your idea is useful because it gives you the equation (after cancelling 5) $$ x^2-y^2-9(x-y)=0. $$ Now use $x^2-y^2=(x+y)(x-y)$ to factor out $x-y$ $$ (x-y)(x+y-9)=0. $$


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