= Solution
For positive edge lengths the <volume> is $V=abc$ and the <surface area> constraint is $2(ab+ac+bc)=S$. Introduce a <Lagrange multiplier> $\mu$:
$$
\mathcal L=abc-\mu\bigl(2ab+2ac+2bc-S\bigr).
$$
The <stationary point> equations are
$$
bc=2\mu(b+c),\qquad ac=2\mu(a+c),\qquad ab=2\mu(a+b).
$$
Multiplying these by $a,b,c$, respectively, and equating their left sides gives $\mu c(a-b)=0$ and $\mu b(a-c)=0$. Because all edge lengths are positive, the equations imply $\mu>0$; hence $a=b=c$. The constraint gives $6a^2=S$.
To establish a global <maximum>, rather than only a <stationary point>, apply the <arithmetic-geometric mean inequality> to $ab,ac,bc$:
$$
\frac S6=\frac{ab+ac+bc}{3}\ge\bigl((ab)(ac)(bc)\bigr)^{1/3}
=(abc)^{2/3}.
$$
Equality holds exactly when $ab=ac=bc$, or $a=b=c$. Therefore
$$
\boxed{V_{\max}=\left(\frac S6\right)^{3/2},\qquad
a=b=c=\sqrt{S/6}.}
$$
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