Let's solve this problem.

We need to find the point on the line $y = x$ that is closest to the point (1, 0). A hint suggests that it is easier to minimize the square of the distance, rather than the distance directly.

Step-by-Step Solution

1. Distance Formula:

   The general formula for the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$

   Here, the first point is $(x, x)$ on the line $y = x$, and the second point is $(1, 0)$. So the distance becomes: $d = \sqrt{(x - 1)^2 + (x - 0)^2} = \sqrt{(x - 1)^2 + x^2}.$

2. Square of the Distance:

   To avoid dealing with square roots, we minimize the square of the distance: $D^2 = (x - 1)^2 + x^2.$ Expand the square: $D^2 = x^2 - 2x + 1 + x^2 = 2x^2 - 2x + 1.$

3. Minimizing the Function:

   Now, we treat $D^2 = 2x^2 - 2x + 1$ as a function of $x$ and minimize it by finding its critical points. Take the derivative with respect to $x$: $\frac{d}{dx}(2x^2 - 2x + 1) = 4x - 2.$ Set the derivative equal to zero to find the critical points: $4x - 2 = 0 \quad \Rightarrow \quad 4x = 2 \quad \Rightarrow \quad x = \frac{1}{2}.$

4. Finding the Corresponding Point:

   Since the point is on the line $y = x$, the corresponding $y$-coordinate is also $y = \frac{1}{2}$. So, the point on the line closest to $(1, 0)$ is: $\left(\frac{1}{2}, \frac{1}{2}\right).$

5. Verification:

   The square of the distance from $\left(\frac{1}{2}, \frac{1}{2}\right)$ to $(1, 0)$ is: $D^2 = \left(\frac{1}{2} - 1\right)^2 + \left(\frac{1}{2} - 0\right)^2 = \left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.$ This confirms our solution is correct.

Final Answer:

The point on the line $y = x$ that is closest to the point $(1, 0)$ is: $\left(\frac{1}{2}, \frac{1}{2}\right).$

Do you want a detailed explanation of any step or further questions?

Related Questions:

1. How do you use calculus to minimize a multivariable function?
2. Why is it easier to minimize the square of the distance instead of the distance?
3. What are critical points, and how do they help solve optimization problems?
4. Can we use geometric intuition to solve this problem without calculus?
5. How do we generalize this technique for different types of curves or lines?

Tip: Always try simplifying distance-based optimization problems by minimizing the square of the distance to avoid working with square roots!