Let’s analyze the problem shown in your image. The task is to identify the critical points of the function $z = x^2 - 2xy + y^2$ and classify them.

Step-by-step Solution:

1. Given function: $z = x^2 - 2xy + y^2$

2. Find the partial derivatives: The critical points occur when both partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ are zero.

   • Partial derivative with respect to $x$: $\frac{\partial z}{\partial x} = 2x - 2y$

   • Partial derivative with respect to $y$: $\frac{\partial z}{\partial y} = -2x + 2y$

3. Set the partial derivatives equal to zero: Solve the system of equations: $2x - 2y = 0 \quad \text{(1)}$ $-2x + 2y = 0 \quad \text{(2)}$

   From equation (1): $x = y$

   Substitute $x = y$ into equation (2): $-2x + 2x = 0$

   This confirms $x = y$ works for all solutions. Hence, the critical point is: $(x, y) = (0, 0)$

4. Classify the critical point: To classify the critical point, compute the second partial derivatives:

   • $\frac{\partial^2 z}{\partial x^2} = 2$
   • $\frac{\partial^2 z}{\partial y^2} = 2$
   • $\frac{\partial^2 z}{\partial x \partial y} = -2$

   Compute the determinant of the Hessian matrix:

   \begin{vmatrix} \frac{\partial^2 z}{\partial x^2} & \frac{\partial^2 z}{\partial x \partial y} \\ \frac{\partial^2 z}{\partial x \partial y} & \frac{\partial^2 z}{\partial y^2} \end{vmatrix} = \begin{vmatrix} 2 & -2 \\ -2 & 2 \end{vmatrix}$$ Compute $$H$$: $$H = (2)(2) - (-2)(-2) = 4 - 4 = 0$$ Since the determinant of the Hessian matrix is zero, the test is inconclusive. Further analysis would be required to classify the critical point.

Summary:

• The critical point is $(0, 0)$.
• The classification is inconclusive using the Hessian determinant.

Let me know if you'd like detailed explanations for any step or further classification using another method!

Questions to Explore:

1. What is the significance of a critical point in multivariable calculus?
2. How does the Hessian determinant help classify critical points?
3. What happens when the determinant of the Hessian matrix is zero?
4. Can this function have a saddle point at $(0, 0)$? How would you determine this?
5. How can eigenvalues of the Hessian matrix help classify critical points?

Tip:

When the Hessian determinant test fails (i.e., $H = 0$), consider analyzing the function's behavior in different directions to classify the critical point.