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I never had seen this exercise, but I'm confused again, I don't know what I have to use.

I have the surface $S=\{(x,y,z)\in \mathbb{R}^3|xy+xz+yz=1,x>0,y>0,z>0\}$, is $S$ regular?. Then, if $S$ is regular, I have to found the higher value of $f$ defined by $f(x,y,z)=xyz$ where $(x,y,z)$ is in $S$.

Can anyone help me?

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There are three projections $S \to \mathbb R^2$, given by $$(x,y,z)\mapsto\begin{cases}(x,y) \\ (x,z) \\ (y,z).\end{cases}$$ For each of these maps, figure out where it is regular. If (for every point $p \in S$, at least one of these three maps is regular at $p$), then $S$ is regular. – Charles Staats Nov 10 '12 at 17:10

Define $F:\mathbb{R}_{x,y,z>0} \to \mathbb{R}$ by $F(x,y,z) = xy + xz + yz.$ Then $dF = (y+z, x+z, x+y)^T$ so $F$ has no critical values, and $1$ is a regular value of $F.$ The preimage of a regular value is a regular surface, so $S= F^{-1}(1)$ is a regular surface.

Then for points on the surface $xy+xz+yz=1$, we have $z = \frac{1-xy}{x+y}$ so we have to maximize $$g(x,y)=\frac{ xy (1-xy)}{x+y}$$ over $\mathbb{R_{x,y>0}}.$

We compute that $g_x = -\frac{y}{(x+y)^2} (x^2+2xy-1)$ and $g_y = -\frac{x}{(x+y)^2} (y^2+2xy-1)$ so $g_x=0$ when $y= \frac{1-x^2}{2x}$ and $g_y=0$ when $x= \frac{1-y^2}{2y}.$ When both these hold we have $1-x^2=2xy=1-y^2$ so $x=y$, and $x^2+2xy-1=0$ leads to $x=y=\frac{1}{\sqrt{3}}$ so the maximum value of $g$ is $\frac{1}{3\sqrt{3}}.$

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To prove that $S$ is regular, prove that the gradient of $xy+xz+yz$ is not zero at all points of $S$.

To maximize $f$ on $S$, use Lagrange multipliers.

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