91Ó°ÊÓ

Constrained optimization is a crucial technique in mathematics for finding the maximum or minimum of a function while adhering to certain restrictions or constraints. In the case of using Lagrange multipliers, we deal with functions with one or more constraints.
This approach allows us to efficiently search for optimal points without needing to repeatedly substitute the constraint directly into the function. Here, we are given a constraint, such as an equation relating the variables, and we aim to optimize another function within this constraint.

• The given function to optimize: \( f(x,y) = xy \)
• The constraint: \( x^2 + y^2 - xy = 9 \)

By introducing the Lagrange multiplier \( \lambda \), we create a new function \( F(x, y, \lambda) \) that incorporates both the original function and the constraint. This new composite function includes the potential to change based on the constraint, making it a powerful tool for optimization.

Critical points are where the first derivative or gradient of a function equals zero, indicating potential maximum, minimum, or saddle points. To find them using Lagrange multipliers, we set partial derivatives of the modified function \( F(x, y, \lambda) \) to zero.
In the exercise, critical points are determined by solving a system of equations arising from these derivatives:

• \( \frac{\partial F}{\partial x} = y - \lambda(2x - y) = 0 \)
• \( \frac{\partial F}{\partial y} = x - \lambda(2y - x) = 0 \)
• \( \frac{\partial F}{\partial \lambda} = x^2 + y^2 - xy - 9 = 0 \)

Solving these equations helps us identify combinations of \(x\), \(y\), and \(\lambda\) that satisfy both the function and constraint conditions. Remember, finding these points doesn't immediately tell you if the point is a maximum or minimum – further analysis is needed to conclude that.

Partial derivatives are essential in understanding how a function like \( F(x, y, \lambda) \) changes with each variable, holding the others constant. They're used to construct the equations that will reveal the critical points.
For a function \( F(x, y, \lambda) \), partial derivatives describe the function's slope along each axis (x, y, \lambda). Using Lagrange multipliers, we took the partial derivatives:

• \( \frac{\partial F}{\partial x} = y - \lambda(2x - y) \)
• \( \frac{\partial F}{\partial y} = x - \lambda(2y - x) \)
• \( \frac{\partial F}{\partial \lambda} = x^2 + y^2 - xy - 9 \)

These derivatives help us set up a system of equations whose solutions are points where the constraints and the goal complement each other perfectly. By setting these derivatives to zero, we find balance points where the maximum or minimum under set constraint conditions is possible.

Once critical points are located, determining their optimality regarding the original function is crucial. This is where we substitute the points back into \( f(x, y) \) to find their values. In our exercise, the critical points were \((3, 3, \frac{1}{2}), (-3, -3, \frac{1}{2}), (3, -3, -\frac{1}{2}), (-3, 3, -\frac{1}{2})\).
Evaluating \( f(x, y) \) at these points gives:

• \( f(3, 3) = 3(3) = 9 \)
• \( f(-3, -3) = (-3)(-3) = 9 \)
• \( f(3, -3) = 3(-3) = -9 \)
• \( f(-3, 3) = (-3)(3) = -9 \)

This reveals the optimal values for the function, indicating that the maximum value is 9 at \((3, 3)\) and \((-3, -3)\), while the minimum value is -9 at \((3, -3)\) and \((-3, 3)\). This concept is practical for finding real-world optimal solutions where conditions or budgets limit the solutions.