# I think my algebra died

• Oct 22nd 2009, 11:45 AM
billym
I think my algebra died
Why am I getting stuck on these questions?

Find and classify the stationary points:

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

$f_x=3(x-y)^2+2xy^2$

$f_y=-3(x-y)^2+2x^2y$

I forget how to find the roots. What to I need to revise? Any links?
• Oct 22nd 2009, 12:20 PM
tonio
Quote:

Originally Posted by billym
Why am I getting stuck on these questions?

Find and classify the stationary points:

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

$f_x=3(x-y)^2+2xy^2$

$f_y=-3(x-y)^2+2x^2y$

I forget how to find the roots. What to I need to revise? Any links?

It is not trivial and requires quite some tricky algebra:

$f_x=0=f_y \Longrightarrow 3(x-y)^2+2xy^2=-3(x-y)^2+2x^2y$ $\Longrightarrow 6(x-y)^2-2xy(x-y)=0 \Longrightarrow$

$\Longrightarrow 3(x-y)\left[2(x-y)-xy\right]=0$ $\Longrightarrow x=y\,,\,\,or \,\,2(x-y)=xy\Longleftrightarrow x-y=\frac{xy}{2}$

If x = y then plugging into $f_x$ we get $x=y=0$, else we plug $x-y=\frac{xy}{2}\,\,in\,\,f_x\,,\,\,f_y$ and:

$f_x=3\left(\frac{xy}{2}\right)^2+2xy^2 \Longrightarrow xy^2 \left(\frac{3}{4}x+2\right)=0$, and again $x = y = 0$ or else $x=-\frac{8}{3}$,
and doing the same with the other partial derivative we get that also $y=-\frac{8}{3}$

Nervertheless these last two values for x,y don't work, as you can check at once, so we're left with only...

Tonio
• Oct 22nd 2009, 01:03 PM
hjortur
Quote:

$
\Longrightarrow 6(x-y)^2-2xy(x-y)=0 \Longrightarrow
$

$
\Longrightarrow 3(x-y)\left[2(x-y)-xy\right]=0
$

There is a mistake there, it is supposed to be

$
\Longrightarrow 2(x-y)\left[3(x-y)-xy\right]=0
$