Question

The optimal value of the constrained optimization problem \[ \text{minimize } 2xy \quad \text{subject to } 9x^2 + 4y^2 \leq 36 \] is _____________. (in integer) 
 

Hint:

For quadratic constraints, use the Lagrange multiplier method to find extrema; symmetry often implies $\pm$ pairs of optimal points.

Answer 1

Correct Answer: -6

Step 1: Write constraint in equality form (at optimum).
\[ 9x^2 + 4y^2 = 36. \]
Step 2: Use Lagrange multiplier method.
Define \[ \mathcal{L} = 2xy + \lambda(9x^2 + 4y^2 - 36). \] Differentiate: \[ \frac{\partial \mathcal{L}}{\partial x} = 2y + 18\lambda x = 0 \quad (1), \] \[ \frac{\partial \mathcal{L}}{\partial y} = 2x + 8\lambda y = 0 \quad (2). \]
Step 3: Eliminate $\lambda$.
From (1): $\lambda = -\dfrac{y}{9x}$, From (2): $\lambda = -\dfrac{x}{4y}$. Equating: \[ \frac{y}{9x} = \frac{x}{4y} \Rightarrow 4y^2 = 9x^2 \Rightarrow y = \pm \frac{3x}{2}. \]
Step 4: Substitute in constraint.
\[ 9x^2 + 4\left(\frac{3x}{2}\right)^2 = 36 \Rightarrow 9x^2 + 9x^2 = 36 \Rightarrow 18x^2 = 36 \Rightarrow x = \pm 1. \] Then $y = \pm \frac{3}{2}$.
Step 5: Compute objective function.
\[ 2xy = 2(1)\left(-\frac{3}{2}\right) = -3, \text{ and } 2(-1)\left(\frac{3}{2}\right) = -3. \] Minimum value occurs at \[ \boxed{2xy = -3.} \]