To find the critical values of the function $f(x, y) = x^2 + y^2$ subject to the constraint $x + y = 1$, we can use the method of Lagrange multipliers. This method allows us to find the extrema of a function subject to a constraint by introducing an auxiliary variable (the Lagrange multiplier).

Steps to Solve Using Lagrange Multipliers:

1. Define the Objective Function and the Constraint:

   • Objective function: $f(x, y) = x^2 + y^2$
   • Constraint: $g(x, y) = x + y - 1 = 0$

2. Set Up the Lagrange Function:

   Define the Lagrange function $\mathcal{L}(x, y, \lambda)$ as: $\mathcal{L}(x, y, \lambda) = f(x, y) - \lambda \cdot g(x, y)$ Substituting the expressions for $f(x, y)$ and $g(x, y)$: $\mathcal{L}(x, y, \lambda) = x^2 + y^2 - \lambda(x + y - 1)$

3. Compute the Partial Derivatives:

   Compute the partial derivatives of $\mathcal{L}(x, y, \lambda)$ with respect to $x$, $y$, and $\lambda$: $\frac{\partial \mathcal{L}}{\partial x} = 2x - \lambda = 0 \quad \Rightarrow \quad 2x = \lambda \quad \text{(1)}$ $\frac{\partial \mathcal{L}}{\partial y} = 2y - \lambda = 0 \quad \Rightarrow \quad 2y = \lambda \quad \text{(2)}$ $\frac{\partial \mathcal{L}}{\partial \lambda} = -(x + y - 1) = 0 \quad \Rightarrow \quad x + y = 1 \quad \text{(3)}$

4. Solve the System of Equations:

   From equations (1) and (2): $2x = \lambda \quad \text{and} \quad 2y = \lambda \quad \Rightarrow \quad 2x = 2y \quad \Rightarrow \quad x = y \quad \text{(4)}$

   Substitute $x = y$ into equation (3): $x + x = 1 \quad \Rightarrow \quad 2x = 1 \quad \Rightarrow \quad x = \frac{1}{2}$

   Using equation (4), $y = x = \frac{1}{2}$.

   Therefore, the critical point is $\left( \frac{1}{2}, \frac{1}{2} \right)$.

5. Determine Whether It's a Minimum or Maximum:

   To determine if this critical point is a minimum or maximum, evaluate the function $f(x, y)$ at the critical point: $f\left( \frac{1}{2}, \frac{1}{2} \right) = \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

   Consider other points on the constraint to check if this value is indeed a minimum or maximum:

   • For $x = 0$, $y = 1$: $f(0, 1) = 0^2 + 1^2 = 1$
   • For $x = 1$, $y = 0$: $f(1, 0) = 1^2 + 0^2 = 1$

   The function value $\frac{1}{2}$ at $\left( \frac{1}{2}, \frac{1}{2} \right)$ is less than the value at other points on the constraint. This indicates that $\left( \frac{1}{2}, \frac{1}{2} \right)$ is a minimum.

Conclusion:

The critical point $\left( \frac{1}{2}, \frac{1}{2} \right)$ is a minimum of the function $f(x, y) = x^2 + y^2$ subject to the constraint $x + y = 1$.

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Would you like more details on any part of the solution or have additional questions?

Here are some related questions:

1. How would the solution change if the constraint were $x + y = 2$?
2. Can Lagrange multipliers be used for functions with more than one constraint?
3. What is the geometric interpretation of Lagrange multipliers?
4. How does the method of Lagrange multipliers compare with other optimization techniques?
5. How would you verify if the critical point is a global minimum or maximum?

Tip: The method of Lagrange multipliers is powerful for constrained optimization problems, especially in situations where direct substitution is difficult. It provides a systematic approach to finding extrema.