We are tasked with finding the maximum value of $f(x, y) = x^2 + y^2$, subject to the constraint $x + y = 1$.

Step 1: Express one variable in terms of the other using the constraint.

From the constraint $x + y = 1$, we can solve for $y$ in terms of $x$: $y = 1 - x$

Step 2: Substitute into the objective function.

Substitute $y = 1 - x$ into $f(x, y) = x^2 + y^2$: $f(x, 1 - x) = x^2 + (1 - x)^2$ Now simplify the expression: $f(x, 1 - x) = x^2 + (1 - 2x + x^2) = x^2 + 1 - 2x + x^2 = 2x^2 - 2x + 1$

Step 3: Find the critical points.

To find the maximum, we take the derivative of $f(x, 1 - x) = 2x^2 - 2x + 1$ with respect to $x$ and set it equal to zero: $\frac{d}{dx}(2x^2 - 2x + 1) = 4x - 2$ Set the derivative equal to zero to find the critical points: $4x - 2 = 0$ $x = \frac{1}{2}$

Step 4: Calculate the corresponding value of $y$.

Using the constraint $x + y = 1$, if $x = \frac{1}{2}$, then: $y = 1 - \frac{1}{2} = \frac{1}{2}$

Step 5: Calculate the maximum value of $f(x, y)$.

Substitute $x = \frac{1}{2}$ and $y = \frac{1}{2}$ into the original function: $f\left( \frac{1}{2}, \frac{1}{2} \right) = \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

Conclusion:

The maximum value of $f(x, y) = x^2 + y^2$ subject to the constraint $x + y = 1$ is $\frac{1}{2}$.

Do you need further details or have any questions?

Here are 5 related questions to expand your understanding:

1. How do you find the minimum value of $f(x, y) = x^2 + y^2$ under the same constraint?
2. What happens to the maximum if the constraint is changed to $x + y = 2$?
3. Can you solve this problem using the method of Lagrange multipliers?
4. How does the function $f(x, y) = x^2 + y^2$ behave for different types of constraints?
5. How would the solution change if $f(x, y)$ was $f(x, y) = x^2 - y^2$ with the same constraint?

Tip: To optimize functions under constraints, it's useful to express one variable in terms of the other, simplifying the problem to one variable.