# Stationary point of Lagrangian

The situation:

Optimise $f(x,y)=x+y^2$

subject to $x^3+3x^2y^2+3y^4=7$.

I write the Lagrangian as $L(x,y,\lambda)=x+y^2-\lambda(x^3+3x^2y^2+3y^4-7)$ and partially differentiate each variable:

$L'_x(x,y,\lambda)=1-3\lambda x^2-6\lambda x y^2$

$L'_y(x,y,\lambda)=2y-6\lambda x^2y-12\lambda y^3$

$L'_\lambda(x,y,\lambda)=-(x^3+3x^2y^2+3y^4-7)$

which gives me the system:

1) $1-3\lambda x^2-6\lambda x y^2=0$

2) $2y-6\lambda x^2y-12\lambda y^3=0 \Rightarrow 1-3\lambda x^2-6\lambda y^2=0$

3) $-x^3-3x^2y^2-3y^4+7=0$

Then I equate equation 1) and 2) which solves for $x=1$. Plugging $x=1$ back into equation 3) gives me $y=1$ or $y=-1$.

The question is, how can the Lagrangian have 3 stationary points in this case? it should be evident that without solving for $\lambda$ that I only get two stationary points:

$(1,1,\lambda)$ and $(1,-1,\lambda)$

Yet, there should be three stationary points, what do I miss?

• You removed a factor of $y$, Was there some reason to exclude $y=0$? Mar 26, 2018 at 20:23
• What do you mean exactly? If I insert $x=1$ into the third formula I eventually get $y^2+y^4 - 2=0$.
– SOMI
Mar 26, 2018 at 20:28
• In equation $2$, you removed a factor of $y$. Why? Mar 26, 2018 at 20:29
• I am fairly new to systems of multivariate equations, I simply thought I could divide equation 2) by $2y$ in order to make the elimination process in the next step easier. Is that fundamentally wrong?
– SOMI
Mar 26, 2018 at 20:33
• Yes, it's wrong unless you know $y\ne 0$. The factor of $y$ yields two cases. In one case, you have $y=0$. In the other case, you have the other factor is zero. By removing the factor of $y$, you lost a case. Mar 26, 2018 at 20:34