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?

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    $\begingroup$ What's the domain of $g$? $\endgroup$
    – JonathanZ
    Commented Dec 28, 2020 at 5:05
  • $\begingroup$ @JonathanZsupportsMonicaC: Hi, the domain of $g$ is $\mathbb{R}$. $\endgroup$
    – Gwyn
    Commented Dec 28, 2020 at 5:30
  • $\begingroup$ Are all the manipulations you have done valid for all $x$ in $\mathbb R$? $\endgroup$
    – JonathanZ
    Commented Dec 28, 2020 at 5:40
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    $\begingroup$ @JonathanZsupportsMonicaC You are right. I was trying to give a different perspective, different from the purely algebraic one. $\endgroup$
    – GReyes
    Commented Dec 28, 2020 at 6:14
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    $\begingroup$ @JonathanZsupportsMonicaC: Thanks for your approach, I agree with you that it is better for learning. Well, $y^2=\sqrt{1-4x^2}\iff y=-\sqrt{1-4x^2}$ or $y=\sqrt{1-4x^2}$ so it must be $1-4x^2 \geq 0 \iff -\frac{1}{2} \leq x \leq \frac{1}{2}$. So this means that I must restrict $g$ to $-\frac{1}{2} \leq x \leq \frac{1}{2}$ and then check by hand the values $x=-\frac{1}{2}$ and $x=\frac{1}{2}$ because derivatives only give information about extrema in $-\frac{1}{2} < x < \frac{1}{2}$? $\endgroup$
    – Gwyn
    Commented Dec 28, 2020 at 6:16

1 Answer 1



$$ \min_{x,y}(\max_{x,y})(x^2+y^2)\ \ \text{s. t.}\ \ \ 4x^2+y^2=1 $$

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.

enter image description here

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 $\frac 14$.


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