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When working with Lagrange multipliers, one of the first steps is computing the gradient vector, which is a crucial part of the process. A gradient vector, represented as \( abla f(x, y) \), shows us the direction of the steepest ascent of a function. For the function \( f(x, y) = x + y \), the gradient is straightforward to compute:

- The partial derivative with respect to \( x \) is 1.
- The partial derivative with respect to \( y \) is also 1.

Thus, the gradient vector \( abla f(x, y) \) is \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \). This vector indicates that the function grows naturally in both the \( x \) and \( y \) directions.See how it contrasts with the gradient of the constraint \( g(x, y) = x^2 - xy + y^2 - 1 \):
- The partial derivative of \( g \) with respect to \( x \) yields \( 2x - y \).
- With respect to \( y \), it gives \( -x + 2y \).

So, the gradient vector for \( g(x, y) \) is \( \begin{bmatrix} 2x - y \ -x + 2y \end{bmatrix} \). Knowing how to compute these gradient vectors is essential for solving the optimization problem using Lagrange multipliers.

Constraint optimization is an essential technique in different fields, artfully balancing a need to maximize or minimize a function while adhering to specific limitations, known as constraints. Here, this technique is illustrated through the function \( f(x, y) = x + y \) with the constraint \( x^2 - xy + y^2 = 1 \).

Lagrange multipliers assist in solving these complex problems by converting them into a system of equations using an auxiliary variable, \( \lambda \) (the Lagrange multiplier). The key to doing this is ensuring that the gradient of the function \( f \) follows the constraint defined by \( g(x, y) \). The relationship can be described as:
\[ abla f(x, y) = \lambda abla g(x, y) \]
This condition leads us to two critical equations matched to the partial derivatives of \( g(x, y) \):

- \( 1 = \lambda (2x - y) \)
- \( 1 = \lambda (-x + 2y) \)

Solving these equations in tandem with the original constraint allows us to locate points where the function's growth, subject to the constraint, is either maximized or minimized. By paralleling gradient vectors, Lagrange multipliers pathfind the nexus where our desired optimal values rest.

Critical points are the outcomes of these calculations, representing values where a function, with a constraint, reaches an extremum – either a maximum or a minimum.

Here's how this works with our function \( f(x, y) = x + y \) and constraint. After setting up the Lagrange equations and solving, we get expressions for \( x \) and \( y \). By expressing \( x \) in terms of \( y \) using:
\[ x = \frac{1 + 4y}{5} \]
And substituting back into our constraint \( g(x, y) = 0 \), we can solve for specific \( y \) values like \( y=\frac{1}{3} \) and \( y=3 \). Each \( y \) value pinpoints a corresponding \( x \) value.

The critical points \((\frac{1}{3}, \frac{1}{3})\) and \((1, 3)\), when evaluated, yield \( f(\frac{1}{3}, \frac{1}{3}) = \frac{2}{3} \) and \( f(1, 3) = 4 \). These results reveal the minimum and maximum values of \( f(x, y) \) under the given constraint.

Locating critical points is a fundamental part of constrained optimization. These points provide insight into how the function behaves and elucidate the optimal solutions within the context of limitations dictated by the constraint.