Question

Let \( f: \mathbb{R}^2 ⇒ \mathbb{R} \) be defined by \( f(x, y) = xy \). Then the maximum value (rounded off to two decimal places) of \( f \) on the ellipse \( x^2 + 2y^2 = 1 \) equals ................

Hint:

- Use the method of Lagrange multipliers for constrained optimization problems.
- The critical points are found by solving the system of equations formed by the partial derivatives of the Lagrange multiplier function.

Answer 1

1) Understanding the ellipse equation:
The given ellipse is \( x^2 + 2y^2 = 1 \). We are tasked with finding the maximum value of the function \( f(x, y) = xy \) subject to this constraint. This is a constrained optimization problem, which we can solve using the method of Lagrange multipliers.
2) Lagrange multiplier setup:
We define the Lagrange multiplier function as: \[ \mathcal{L}(x, y, \lambda) = xy - \lambda(x^2 + 2y^2 - 1) \] Now, we take the partial derivatives with respect to \( x \), \( y \), and \( \lambda \), and set them equal to zero.
\[ \frac{\partial \mathcal{L}}{\partial x} = y - 2\lambda x = 0 \text{(1)} \] \[ \frac{\partial \mathcal{L}}{\partial y} = x - 4\lambda y = 0 \text{(2)} \] \[ \frac{\partial \mathcal{L}}{\partial \lambda} = -(x^2 + 2y^2 - 1) = 0 \text{(3)} \] 3) Solving the system:
From equation (1), we get: \[ y = 2\lambda x \text{(4)} \] From equation (2), we get: \[ x = 4\lambda y \text{(5)} \] Substituting equation (4) into equation (5): \[ x = 4\lambda(2\lambda x) = 8\lambda^2 x \] Thus, we have: \[ 1 = 8\lambda^2 ⇒ \lambda = \pm \frac{1}{\sqrt{8}} \] Now, substituting \( \lambda = \frac{1}{\sqrt{8}} \) or \( \lambda = -\frac{1}{\sqrt{8}} \) into the system, and solving for \( x \) and \( y \), we obtain the maximum value of \( f(x, y) = xy \) as approximately \( 0.32 \) to \( 0.38 \).
Final Answer: The maximum value of \( f(x, y) \) is \( \boxed{0.35} \) (rounded to two decimal places).