Let $z =2 x^2+x^2y+2xy+xy^2+\frac{5}{2}y^2+\frac{1}{6}y^3$. Find and classify, the stationary points of $z$.

Question

I know $\frac{dz}{dx}=4x+2xy+2y+y^2$ and $\frac{dz}{dy}=x^2+2x+2xy+5y+\frac{1}{2}y^2$ and I know that to find the stationary point it is the point at which they both equal 0 right?

Just wandering what to do next?

Make $4x+2xy+2y+y^2=0$ and $x^2+2x+2xy+5y+\frac{1}{2}y^2=0$ then how do I solve simultaneously? And what comes after that?

Sorry if this is a really basic question. Thanks for your help!

Answer 1

(I am assuming the first term is $2x^2$.)

Since $4x+2xy+2y+y^2=0$ $\;\;$ [1]

and $x^2+2x+2xy+5y+\frac{1}{2}y^2=0$ $\;\;$ [2], $\;\;$subtracting [1] from [2] gives

$\;\;\;\;\;x^2-2x+3y=\frac{1}{2}y^2$ $\;\;$ [3], $\;\;$and substituting back into [2] gives

$x^2+2x+2xy+5y+(x^2-2x+3y)=0$,$\;\;$ so $\;2x^2+2xy+8y=0$ and therefore

$\;\;x^2+xy+4y=0$ and so $\displaystyle y=-\frac{x^2}{x+4}$. $\;\;$Substituting into [3] gives

$\displaystyle x^2-2x-\frac{3x^2}{x+4}=\frac{x^4}{2(x+4)^2}$, and multiplying by $2(x+4)^2$ gives

$2(x^2-2x)(x+4)^2-6x^2(x+4)=x^4$. Multiplying out yields

$2(x^4+6x^3-24x^2-64x)=0$, $\;\;$so $\;\;2x(x^3+6x^2-24x-64)=0$.

Thus $x\left((x^3-64)+6x(x-4)\right)=0$, so $x\left((x-4)(x^2+4x+16)+6x(x-4)\right)=0$ gives

$x(x-4)(x^2+10x+16)=0$ and so $\;\;x(x-4)(x+2)(x+8)=0$.

Therefore the critical points are $(0,0),\; (4,-2), \;(-2,-2),\;(-8,16)$.

Answer 2

$$4x+2xy+2y+y^2=0\\ x^2+2x+2xy+5y+\frac12y^2=0$$ Use the first equation to eliminate $x$. $$\frac{(2y+y^2)^2}{4(2+y)^2}-\frac{(2+2y)(2y+y^2)}{4+2y}+5y+\frac12y^2=0$$
There seem to be four points where $y=0$ Then feed each one back into the first equation to find $x$.
Or, the first line factors as $(2x+y)(2+y)$. Substitute either $y=-2$ or $y=-2x$ in the second equation.