I am trying to figure out the critical points of a multivariable function $\ f(x,y)=(x-2)(y-2)(x+y-2)$

Question

After multiplying it out I get: $\ f(x,y)=x^2y+xy^2-6xy-2x^2-2y^2+8x+8y-8$

Where I would take the partial derivative

with respect to x and set to zero: $\ y^2+2xy-6y-4x+8=0$,

with respect to y and set to zero: $\ x^2+2xy-6x-4y+8=0$

This is where I am stuck and unable to proceed. Any tips would be helpful.

Answer 1

Since both $x^2-6x-4y$ and $y^2-6y-4x$ are equal to $-2xy-8$, you have $x^2-6x-4y=y^2-6y-4x$, or $x^2-y^2=2(x-y)$. Therefore, $x=y$ or $x+y=2$.

If $x=y$, your first equation becomes $3y^2-10y+8=0$, which has two solutions: $\frac43$ and $2$. So, this provides two critical points: $\left(\frac43,\frac43\right)$ and $(2,2)$.

And, if $x=2-y$, your first equation becomes $2y-y^2=0$, which has two solutions: $0$ and $2$. So, this provides another two solutions: $(2,0)$, and $(0,2)$.

Answer 2

Let me try to "cheat" a little: introduce the variables $$ u = x - 2, \; v = y - 2. $$ Then the function becomes $$ F(u, v) = f(x, y) = u v (u + v + 2). $$ The partial derivatives are: $$ F_{u} = v ( (u + v + 2) + u) = v ( 2u + v + 2), \quad F_{v} = u ( (u + v + 2) + v) = u ( u + 2v + 2). $$ Equating $F_{u}$ to zero gives: $$ v = 0 \mbox{ or } v = - 2u - 2. $$ Analogously, equating $F_{v}$ to zero gives: $$ u = 0 \mbox{ or } u = - 2v - 2; $$ At the outset, this seems to give four possibilities. Three of them are: $$ (u = 0, v = 0); \quad (u = -2, v = 0); \quad (u = 0, v = -2). $$ The last one is the intersection of the lines $$ u = - 2v - 2, \quad v = - 2u - 2. $$ Having found the critical points in the $uv$-coordinates, one can convert to the $xy$-coordinates.

Answer 3

Just don't multiply out. The partial derivative with respect to $x$ (note that $(y-2)$ is constant in this case) using the product rule for $(x-2)(x+y-2)$ is $$f_x(x,y) =(y-2)(x+y-2+x-2).$$ Similarly, $$f_y(x,y)=(x-2)(x+y-2+y-2).$$ From here it's a piece of cake.