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I was reading about the problem of maximizing $x^2+y^2+z^2$ on the intersection of the two surfaces $xyz=1$ and $x^2 + y^2 + 2z^2 = 4$. The author wrote that $\nabla F=a \nabla g+b \nabla h$ (for $a$ and $b$ arbitrary numbers and $g$ and $h$ constraints, which are surfaces in this case) then he wrote that $$ <2x, 2y, 2z> = a <yz, xz, xy> + b <2x, 2y, 4z> $$ and from here he said that we have five equations: $$2x = ayz + 2bx$$ $$2y = axz + 2by$$ $$2z = axy + 4bz$$ $$xyz = 1$$ $$x^2 + z = 1$$ The first four are clear but where does the fifth equation come from? Also what is a good way to solve such problems by this method? Thanks everybody, please explain to me.

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Isn't the last equation supposted to be $x^2 + y^2 + 2z^2 = 4$? Perhaps you could provide details about the text you're studying. –  Martin Sleziak Jul 31 '11 at 8:31
The solution looks wrong to me. It looks like the poster changed the problem and forgot to change the solution. As for your last question, this is the standard way to solve such problems. –  TonyK Jul 31 '11 at 8:51
If you have $n$ variables and $r$ constraints then this method leads to a system of $n+r$ equations which in general is highly nonlinear. There is no "simple method" to solve such systems. –  Christian Blatter Jul 31 '11 at 13:51
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1 Answer

up vote 2 down vote accepted

Here is a (hopefully) correct solution of the original problem. Let $S\subset{\mathbb R}^3$ be the set defined by the constraints. The set $S$ consists of 4 smooth loops: The equation $x^2+y^2+2z^2=4$ defines an ellipsoid $E$ whereas $xyz=1$ defines 4 infinite "cups" which intersect $E$ in 4 curves $\gamma_i$.

I replace your fifth equation by the original constraint $x^2+y^2+2z^2=4$.

Multiplying your first equation by $y$, the second by $-x$ and adding the two gives $$0\ =\ a(y^2-x^2) z\ .$$ Now $z\ne 0$ on $S$; furthermore if $a=0$ then the first three equations would imply $(x,y,z)=b (x, y, 2z)$. This implies $x=0$ or $b=1$, whence $z=0$, which both are forbidden on $S$. Therefore necessarily $x=y$ or $x=-y$. For symmetry reasons it is enough to look at the first octant in $(x,y,z)$-space, i.e. at the case $x=y$. In this case the constraints reduce to $$x^2 z=1\ ,\qquad x^2+z^2=2\ .$$ Eliminating $z$ leads to $$x^2+{1\over x^4}=2\quad {\rm or}\quad x^6-2x^4+1=(x^2-1)(x^4-x^2-1)=0\ .$$ The last equation has the real solutions $x_1= 1$ and $x_2=\sqrt{(1+\sqrt{5})/2}$. Therefore on $\gamma_1:=S\cap{\rm \{first\ octant}\}$ we have the two conditionally stationary points $P_1:=(1,1,1)$ and $P_2=(x_2,x_2,2-x_2^2)$ .

The function to maximize is $f(x,y,z):=x^2+y^2+z^2$. The computation gives $$f(P_1)= 3\ ,\qquad f(P_2)=2{1+\sqrt{5}\over2}+{7-3\sqrt{5}\over2}={9\over2}-{\sqrt{5}\over2}\doteq 3.382\ , $$ therefore the $\max$ is $3.382$, and $3$ is in fact the $\min$ (which has to exist by general principles).

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