The problem asks to find all points where the minimum rate of increase of the function $f(x, y) = x^2 + y^2 - 2x - 4y$ is in the direction of $\mathbf{i} + \mathbf{j}$.

Steps to Solve:

1. Gradient of the function: The direction of the steepest increase of a function is given by its gradient, $\nabla f(x, y)$. To find $\nabla f(x, y)$, we calculate the partial derivatives of $f(x, y)$ with respect to $x$ and $y$: $\frac{\partial f}{\partial x} = 2x - 2, \quad \frac{\partial f}{\partial y} = 2y - 4$ Thus, the gradient is: $\nabla f(x, y) = (2x - 2) \mathbf{i} + (2y - 4) \mathbf{j}$

2. Direction of the gradient: The direction in which we are interested is $\mathbf{i} + \mathbf{j}$. This direction vector is: $\mathbf{i} + \mathbf{j} = (1, 1)$

3. Minimum rate of increase: The minimum rate of increase happens when the gradient $\nabla f(x, y)$ is perpendicular to the direction vector $\mathbf{i} + \mathbf{j}$. Two vectors are perpendicular when their dot product is zero. Therefore, we want: $(2x - 2)(1) + (2y - 4)(1) = 0$ Simplifying: $2x - 2 + 2y - 4 = 0$ $2x + 2y - 6 = 0 \quad \Rightarrow \quad x + y = 3$

4. Conclusion: The points where the minimum rate of increase of $f(x, y)$ is in the direction of $\mathbf{i} + \mathbf{j}$ lie on the line $x + y = 3$.

Would you like a more detailed breakdown of any step? Here are some related questions to explore:

1. What is the significance of the gradient in finding the direction of maximum increase?
2. Can the rate of increase ever be negative, and how would that affect this problem?
3. How can we extend this analysis to functions of three variables?
4. What would happen if the direction vector were $\mathbf{i} - \mathbf{j}$ instead?
5. How does the dot product help in identifying perpendicular vectors?

Tip: When finding minimum or maximum rates of change, always consider the gradient vector and its relationship with the desired direction.