To find the critical points of the function $f(x, y) = x^2 + y^2 - 4x - 6y + 5$, we need to follow these steps:

1. Find the first partial derivatives of $f(x, y)$:

We need to compute the partial derivatives with respect to $x$ and $y$, and then set them equal to 0 to find the critical points.

• Partial derivative with respect to $x$: $\frac{\partial f}{\partial x} = 2x - 4$
• Partial derivative with respect to $y$: $\frac{\partial f}{\partial y} = 2y - 6$

2. Set both partial derivatives equal to 0 to find the critical points:

We set both equations equal to zero and solve for $x$ and $y$.

• Set $\frac{\partial f}{\partial x} = 0$: $2x - 4 = 0 \quad \Rightarrow \quad x = 2$
• Set $\frac{\partial f}{\partial y} = 0$: $2y - 6 = 0 \quad \Rightarrow \quad y = 3$

So, the critical point is $(x, y) = (2, 3)$.

3. Classify the critical point using the second derivative test:

To classify the critical point, we need to compute the second partial derivatives of $f(x, y)$:

• Second partial derivative with respect to $x$: $f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2$
• Second partial derivative with respect to $y$: $f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2$
• Mixed partial derivative: $f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0$

The discriminant $D$ is given by: $D = f_{xx} f_{yy} - (f_{xy})^2$ Substitute the values: $D = (2)(2) - (0)^2 = 4$

Since $D > 0$ and $f_{xx} > 0$, the critical point $(2, 3)$ is a local minimum.

Conclusion:

The function $f(x, y) = x^2 + y^2 - 4x - 6y + 5$ has a critical point at $(2, 3)$, and it is a local minimum.

Would you like more details on this solution or have any further questions? Here are some related questions for further exploration:

1. How would the classification change if $D < 0$?
2. How do we find the global minimum or maximum of a function in two variables?
3. What happens if $f_{xx}$ and $f_{yy}$ have opposite signs?
4. How does the second derivative test for functions of two variables relate to concavity?
5. How do you apply this to functions with more than two variables?

Tip: For functions of two variables, always check if the second partial derivatives are continuous and verify the discriminant for classification.