# 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?

• What's the domain of $g$? Commented Dec 28, 2020 at 5:05
• @JonathanZsupportsMonicaC: Hi, the domain of $g$ is $\mathbb{R}$.
– Gwyn
Commented Dec 28, 2020 at 5:30
• Are all the manipulations you have done valid for all $x$ in $\mathbb R$? Commented Dec 28, 2020 at 5:40
• @JonathanZsupportsMonicaC You are right. I was trying to give a different perspective, different from the purely algebraic one. Commented Dec 28, 2020 at 6:14
• @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 $-\frac{1}{2} < x < \frac{1}{2}$?
– Gwyn
Commented Dec 28, 2020 at 6:16

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 $$\frac 14$$.