This problem asks us to use the Kuhn-Tucker (also known as the Karush-Kuhn-Tucker, KKT) conditions to find the maxima of $f(x, y) = 2x + 3y$ subject to the constraint $g(x, y) = x^2 + y^2 \leq 2$.

Step-by-Step Solution:

1. Define the Lagrangian:

The Lagrangian $\mathcal{L}(x, y, \lambda)$ is given by: $\mathcal{L}(x, y, \lambda) = f(x, y) + \lambda g(x, y)$ where $f(x, y) = 2x + 3y$ and $g(x, y) = x^2 + y^2 - 2$ (we subtract 2 to express the constraint as $g(x, y) \leq 0$).

Thus, the Lagrangian is: $\mathcal{L}(x, y, \lambda) = 2x + 3y + \lambda (x^2 + y^2 - 2)$

2. Kuhn-Tucker Conditions:

We need to satisfy the following conditions:

• Stationarity: The gradients of the Lagrangian with respect to $x$, $y$, and $\lambda$ must vanish.
• Primal Feasibility: The constraint $g(x, y) \leq 0$ must hold.
• Dual Feasibility: $\lambda \geq 0$.
• Complementary Slackness: $\lambda g(x, y) = 0$.

Stationarity Conditions:

The partial derivatives of $\mathcal{L}(x, y, \lambda)$ are: $\frac{\partial \mathcal{L}}{\partial x} = 2 + 2\lambda x = 0 \quad \Rightarrow \quad \lambda x = -1$ $\frac{\partial \mathcal{L}}{\partial y} = 3 + 2\lambda y = 0 \quad \Rightarrow \quad \lambda y = -\frac{3}{2}$ $\frac{\partial \mathcal{L}}{\partial \lambda} = x^2 + y^2 - 2 = 0$

Solving the Stationarity Equations:

From $\lambda x = -1$, we get: $x = -\frac{1}{\lambda}$ From $\lambda y = -\frac{3}{2}$, we get: $y = -\frac{3}{2\lambda}$ Now, substitute these into the constraint equation $x^2 + y^2 = 2$: $\left(-\frac{1}{\lambda}\right)^2 + \left(-\frac{3}{2\lambda}\right)^2 = 2$ $\frac{1}{\lambda^2} + \frac{9}{4\lambda^2} = 2$ $\frac{1 + \frac{9}{4}}{\lambda^2} = 2 \quad \Rightarrow \quad \frac{13}{4\lambda^2} = 2$ $\lambda^2 = \frac{13}{8} \quad \Rightarrow \quad \lambda = \pm \sqrt{\frac{13}{8}} = \pm \frac{\sqrt{13}}{2\sqrt{2}}$

Since $\lambda \geq 0$ (from the dual feasibility condition), we choose the positive root: $\lambda = \frac{\sqrt{13}}{2\sqrt{2}}$

Finding $x$ and $y$:

Substitute $\lambda$ back into the expressions for $x$ and $y$: $x = -\frac{1}{\frac{\sqrt{13}}{2\sqrt{2}}} = -\frac{2\sqrt{2}}{\sqrt{13}}, \quad y = -\frac{3}{2\left(\frac{\sqrt{13}}{2\sqrt{2}}\right)} = -\frac{3\sqrt{2}}{\sqrt{13}}$

Thus, the solution is: $x = -\frac{2\sqrt{2}}{\sqrt{13}}, \quad y = -\frac{3\sqrt{2}}{\sqrt{13}}$

3. Finding the Maximum:

Finally, evaluate $f(x, y)$ at this point: $f(x, y) = 2x + 3y = 2\left(-\frac{2\sqrt{2}}{\sqrt{13}}\right) + 3\left(-\frac{3\sqrt{2}}{\sqrt{13}}\right)$ $f(x, y) = -\frac{4\sqrt{2}}{\sqrt{13}} - \frac{9\sqrt{2}}{\sqrt{13}} = -\frac{13\sqrt{2}}{\sqrt{13}} = -\sqrt{13}$

Thus, the maximum value of $f(x, y)$ subject to the constraint is $-\sqrt{13}$.

Would you like more details or have any questions on the steps?

Here are some related questions you can explore:

1. How do the Kuhn-Tucker conditions relate to Lagrange multipliers?
2. Can we have multiple maxima if the constraint region changes?
3. How would the result differ if the objective function had higher degrees of $x$ and $y$?
4. How do we interpret the dual variable $\lambda$ in this context?
5. Could the problem be approached using numerical optimization methods?

Tip: Always check the complementary slackness condition to verify that the solution satisfies all conditions of the KKT theorem.