Mistake when substituting constraint $4x^2+y^2=1$ into a function $f(x,y)=x^2+y^2$ in extrema problem Consider the function $f(x,y)=x^2+y^2$ under the constraint $4x^2+y^2=1$. The extrema of $f$ under that constraint can be easily found with Lagrange multipliers, and they are attained for $(0,1)$, $(0,-1)$ for maximum and $(1/2,0)$, $(-1/2,0)$ for minimum; however, if we isolate $y^2=1-4x^2$ and we substitute it in the function, we get a wrong result (in particular, the one variable function $g(x)=1-3x^2$ obtained has only a maximum for $x=0$ and Weierstrass theorem assures that the are both maximum and minimum for $f$). Can someone explain me why this fails?

Solving
$$
\min_{x,y}(\max_{x,y})(x^2+y^2)\ \ \text{s. t.}\ \ \ 4x^2+y^2=1
$$
using the Lagrange multipliers method, reduces to determine the stationary points for
$$
\nabla(x^2+y^2)+\lambda\nabla(4x^2+y^2-1)=0
$$
by solving for $x,y,\lambda$
$$
\cases{
2x+8x\lambda=0\\
2y+2y\lambda=0\\
4x^2+y^2-1=0
}
$$
giving as solutions the four tangency points between the level curves for $z=x^2+y^2$ and the ellipse $4x^2+y^2-1=0$. Those points can be depicted in the attached plot.

Now by making the substitution $y = 1-4x^2$ into $z = x^2+y^2$ giving $z=1-3x^2$ we are searching extrema along the $x$ axis at $x^*=0$ with value $1$ and the corresponding $y^*$ we obtain from the restriction $4(x^*)+(y^*)^2=1$.   Also with the substitution $x^2=\frac{1-y^2}{4}$ we are searching extrema along the $y$ axis with value $\frac 14$.
