Constrained Optimization and Lagrange Multipliers Questions and Answers

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.

Answer: B

S2 is TRUE and S1 is FALSE.

S1: \(f(x_{1}, x_{2}) = (x_{1} - x_{2})^{2}\). Then \(f^{-1}(\{0\})\) is the whole line \(x_{1} = x_{2}\), which is unbounded, so the preimage of the compact set {0} is not compact.

S2: g is a positive definite quadratic form, since its matrix \(\begin{bmatrix} 2 & -1/2 \\ -1/2 & 2 \end{bmatrix}\) has eigenvalues 1.5 and 2.5. Hence \(g(x) \ge 1.5\|x\|^{2}\), so the preimage of a bounded set is bounded; it is also closed because g is continuous. Closed and bounded in \(\mathbb{R}^{2}\) means compact, so S2 holds.