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In this post, the OP wants to minimize $$ f(x,y,z) = x^2+y^2+z^2$$ subject to $$ 4x^2+2y^2+z^2 = 4 $$

Different methods yield $(1,0,0)$ as a minimizer, with $\min{f(x,y,z)}=1$.

Given that $z^2 = 4-4x^2-2y^2$, isn't the problem equivalent to minimizing $$ f(x,y,z(x,y))=4-3x^2-y^2\; ? $$

This seems logical to me, but this function is unbounded below, and has no minimizer. This makes no sense, as $f(x,y,z)\ge 0$.

Can someone find the catch?

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Kuifje
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  • Where's the bounding condition in the second "equivalent" problem? – snowfall512 Mar 07 '17 at 19:59
  • That's what I'm confused with, I think there is no longer a bounding condition, but this is obviously wrong? – Kuifje Mar 07 '17 at 20:10
  • It's wrong simply because there is no bounding condition! – snowfall512 Mar 07 '17 at 20:13
  • What would the bounding condition be in terms of $x$ and $y$? – Kuifje Mar 07 '17 at 20:15
  • You can't make one. It's just not doable. – snowfall512 Mar 07 '17 at 20:24
  • I believe you can. For example, if you minimize a given function subject to $x^2+y^2 = R$, you can solve the problem with $x=R\cos t$ and $y = R \sin t$ and forget about the constraint $x^2+y^2 = R$. This is what we are doing here, but the constraint is $4x^2+2y^2+z^2=4$. – Kuifje Mar 07 '17 at 20:28
  • That requires spherical coordinates, which is 3-variable. The main flaw in trying to set up a constraint is that one of your variables has already been set, and this is 3-space. So you only have two variables available, but a constraint would require 3. – snowfall512 Mar 07 '17 at 20:39
  • I disagree. I will post an answer that a user wrote on the initial post (with the appropriate bounding condition). – Kuifje Mar 07 '17 at 20:55

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In the initial post, @user35734 has found the catch:

Since $z^2$ cannot be negative, the two forms are equivalent only if the following bounding condition is added: $$ 4x^2+2y^2 \le 4 $$

This is an elliptic domain, and it is easy to see that the minimum will be reached on its border, therefore we can replace variables $x$ and $y$ by $\cos t$ and ${\sqrt{2}} \sin t$, respectively. The problem then boils down to minimize $$ f(x(t),y(t))=4-3\cos^2t - 2\sin^2 t $$ Standard calculus techniques yield $$ \min f = 1 $$

Kuifje
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