Discussion
The maximum value of the function \(f(x, y, z) = xyz\) subject to the constraint \(xy + yz + zx - a = 0, a \gt 0\) is
Answer: B
The maximum is \((a/3)^{3/2}\).
Use Lagrange multipliers with \(g = xy + yz + zx - a\): \(\nabla(xyz) = \lambda\nabla g\) gives
\[ \begin{aligned} yz &= \lambda(y + z) \\ zx &= \lambda(z + x) \\ xy &= \lambda(x + y) \end{aligned} \]
For positive variables these force \(x = y = z\). The constraint then gives \(3x^{2} = a\), so \(x = \sqrt{a/3}\) and
\[ xyz = \left(\sqrt{a/3}\right)^{3} = (a/3)^{3/2}. \]
The same bound follows from AM-GM: \(\frac{a}{3} = \frac{xy + yz + zx}{3} \ge (xyz)^{2/3}\). \(a^{3/2}\) drops the factor 3, \((3/a)^{3/2}\) inverts it, and \((3a/2)^{3/2}\) exceeds the AM-GM bound.
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