Solving an optimization problem in multiple variables, I had to examine a function $$ f(x,y,z)=x^2+2yz $$ defined on a ball $$ \{ (x,y,z)\enspace|\enspace x^2+y^2+z^2\leq1 \}. $$ The boundary is then described by the sphere $x^2+y^2+z^2=1$. According to my textbook, one can search for maxima on the boundary by rewriting $x^2+y^2+z^2=1$ as $x^2=1-y^2-z^2$, then substituting this into $f$ to obtain $$ f(x,y,z)=1-y^2-z^2+2yz,\quad y^2+z^2\leq1 $$ Now what I don't understand is, how come we need to specify that $y^2+z^2\leq1$? The unit sphere is described by $x^2+y^2+z^2=1$ with no need to specify any restrictions on $y$ or $z$, but once we rewrite this as $x^2=1-y^2-z^2$ and plug it into the function, suddenly it becomes necessary to specify that $y^2+z^2\leq1$. Why?
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Because $0 \leq x^2 = 1-y^2-z^2$.
Siminore
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But if $y^2+z^2\leq1$ is already implied by $x^2+y^2+z^2=1$, why do I have to write it explicitly? – Andrea Apr 04 '14 at 09:58
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Because you "forget" the $x$ variable. – Siminore Apr 04 '14 at 10:52
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When you write $$x^2+y^2+z^2=1$$
The set of solutions implicitly has the restriction that $y^2+z^2\le1$ because if this condition is violated there are no solutions in $\mathbb R$.
$$y^2+z^2\gt1$$
$$x^2+y^2+z^2\gt1+x^2$$
$$1\gt1+x^2\implies x^2\lt0\implies x\in \phi$$
Guy
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Right, but if it's implicit from the beginning, why do I have to write it explicitly later? Shouldn't it still be implicit? – Andrea Apr 04 '14 at 09:53
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1@Andreas In $f(x,y,z)$ there is no "$\color{red}{=}$". It is simply an expression. You need to point out, that the inputs into $f$ come from an "$\color{red}{=}$"(which is constrained by some inequalities) – Guy Apr 04 '14 at 10:01
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To make myself clearer, $f(x,y,z)$ is defined for all $x,y,z$, but you are only using that definition on the points on the sphere. $f(100,100,100)$ has a very well defined value, but that value is never obtained for the points on the sphere. – Guy Apr 04 '14 at 10:02
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$f$, clearly doesn't have a global maxima, but there is a maximum value for points within or on the sphere. – Guy Apr 04 '14 at 10:03
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Okay, I think I understand. I assumed that, because we inserted $x^2+y^2+z^2=1$ into the function, the function would implicitly contain the equality $x^2+y^2+z^2=1$. I see now that it doesn't. – Andrea Apr 04 '14 at 10:06
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@Andreas yes you're right. It doesn't contain the equality. You need to explicitly enforce that yourself. – Guy Apr 04 '14 at 10:08