Given that $f(x, y)$ is a twice continuously differentiable function and $f$ attains a local maximum at the point $(a, b)$, where the first partial derivatives $\frac{\partial f}{\partial x}(a, b) = 0$ and $\frac{\partial f}{\partial y}(a, b) = 0$, we are interested in the behavior of the second partial derivatives to determine which of the provided statements is true.

Step 1: Analyze the second derivative test for functions of two variables

For a function $f(x, y)$ of two variables, the second derivative test involves the Hessian matrix $H$ at the critical point $(a, b)$:

\frac{\partial^2 f}{\partial x^2}(a, b) & \frac{\partial^2 f}{\partial x \partial y}(a, b) \\ \frac{\partial^2 f}{\partial y \partial x}(a, b) & \frac{\partial^2 f}{\partial y^2}(a, b) \end{pmatrix}$$ The discriminant $$D$$ is given by: $$D = \frac{\partial^2 f}{\partial x^2}(a, b) \cdot \frac{\partial^2 f}{\partial y^2}(a, b) - \left(\frac{\partial^2 f}{\partial x \partial y}(a, b)\right)^2$$ ### Step 2: Applying the second derivative test - **Case 1: Local Maximum** - If $$D > 0$$ and $$\frac{\partial^2 f}{\partial x^2}(a, b) < 0$$ and $$\frac{\partial^2 f}{\partial y^2}(a, b) < 0$$, then $$f$$ has a local maximum at $$(a, b)$$. - **Case 2: Local Minimum** - If $$D > 0$$ and $$\frac{\partial^2 f}{\partial x^2}(a, b) > 0$$ and $$\frac{\partial^2 f}{\partial y^2}(a, b) > 0$$, then $$f$$ has a local minimum at $$(a, b)$$. - **Case 3: Saddle Point** - If $$D < 0$$, then $$(a, b)$$ is a saddle point, regardless of the signs of $$\frac{\partial^2 f}{\partial x^2}(a, b)$$ and $$\frac{\partial^2 f}{\partial y^2}(a, b)$$. - **Case 4: Inconclusive** - If $$D = 0$$, the test is inconclusive. ### Step 3: Identifying the correct statement Given that $$f$$ attains a local maximum at $$(a, b)$$, we must have: - $$\frac{\partial^2 f}{\partial x^2}(a, b) < 0$$ - $$\frac{\partial^2 f}{\partial y^2}(a, b) < 0$$ - $$D > 0$$, which means: $$\frac{\partial^2 f}{\partial x^2}(a, b) \cdot \frac{\partial^2 f}{\partial y^2}(a, b) > \left(\frac{\partial^2 f}{\partial x \partial y}(a, b)\right)^2$$ This corresponds to **option (A)**: $$(A) \,\, \frac{\partial^2 f}{\partial x^2}(a, b) < 0, \, \frac{\partial^2 f}{\partial y^2}(a, b) < 0, \, \frac{\partial^2 f}{\partial x^2}(a, b) \cdot \frac{\partial^2 f}{\partial y^2}(a, b) - \left(\frac{\partial^2 f}{\partial x \partial y}(a, b)\right)^2 > 0$$ **Answer**: The correct statement is **(A)**. --- Would you like to explore any additional details on this topic? Here are some related questions: 1. How does the Hessian matrix help in determining the nature of critical points? 2. Can you explain why $$D > 0$$ indicates a local extremum? 3. What are the implications if $$D = 0$$? 4. How does the second derivative test differ in one-dimensional and multi-dimensional contexts? 5. What is the significance of the mixed partial derivative in the Hessian matrix? 6. Can you give an example where $$D < 0$$ and explain why it results in a saddle point? 7. How does this analysis change for functions of three variables? 8. What are some practical applications of the second derivative test in optimization problems? **Tip**: Always verify the sign of the second derivative discriminant $$D$$ when performing the second derivative test to correctly classify the critical points.