Lagrange's multiplier method is a technique in calculus used to find the local maxima and minima of a function subject to one or more constraints. The method introduces a new variable (called a Lagrange multiplier) for each constraint and transforms the problem into one where the gradients of the objective function and the constraint(s) are parallel. This allows for the optimization of the function while satisfying the constraints.

Main Idea

Suppose you have a function $f(x, y)$ that you want to maximize or minimize, subject to a constraint $g(x, y) = 0$. The main idea behind Lagrange multipliers is to solve the following system of equations:

$\nabla f(x, y) = \lambda \nabla g(x, y)$ $g(x, y) = 0$

Here, $\nabla f(x, y)$ is the gradient of the objective function $f(x, y)$, $\nabla g(x, y)$ is the gradient of the constraint $g(x, y)$, and $\lambda$ is the Lagrange multiplier. The first equation says that the gradients of the objective function and the constraint are parallel at the optimal points, and the second equation ensures the constraint is satisfied.

Example: Maximizing $f(x, y) = x + y$ subject to the constraint $x^2 + y^2 = 25$

1. Objective function: $f(x, y) = x + y$
2. Constraint: $g(x, y) = x^2 + y^2 - 25 = 0$

We want to maximize $f(x, y)$ subject to the constraint $g(x, y) = 0$.

Step 1: Set up the Lagrange multiplier system.

We need to solve:

$\nabla f(x, y) = \lambda \nabla g(x, y)$ $g(x, y) = 0$

First, compute the gradients:

• $\nabla f(x, y) = (1, 1)$
• $\nabla g(x, y) = (2x, 2y)$

Now, set up the system:

$(1, 1) = \lambda (2x, 2y)$ $x^2 + y^2 = 25$

Step 2: Solve the system.

From the first equation, we have:

$1 = 2\lambda x \quad \text{and} \quad 1 = 2\lambda y$

This gives:

$\lambda = \frac{1}{2x} \quad \text{and} \quad \lambda = \frac{1}{2y}$

Equating the two expressions for $\lambda$, we get:

$\frac{1}{2x} = \frac{1}{2y}$

Thus, $x = y$.

Step 3: Substitute into the constraint.

Substitute $x = y$ into the constraint equation:

$x^2 + x^2 = 25$ $2x^2 = 25$ $x^2 = \frac{25}{2}$ $x = \pm \frac{5}{\sqrt{2}} \quad \text{and} \quad y = \pm \frac{5}{\sqrt{2}}$

Step 4: Find the maximum.

Now substitute these values of $x$ and $y$ back into the objective function:

$f(x, y) = x + y = \pm \frac{5}{\sqrt{2}} + \pm \frac{5}{\sqrt{2}} = \pm \frac{10}{\sqrt{2}} = \pm 5\sqrt{2}$

Thus, the maximum value is $5\sqrt{2}$ when $x = y = \frac{5}{\sqrt{2}}$, and the minimum value is $-5\sqrt{2}$ when $x = y = -\frac{5}{\sqrt{2}}$.

Summary

Lagrange's multiplier method finds the maximum and minimum values of a function subject to a constraint by setting up a system where the gradients of the objective function and the constraint are proportional. The multiplier $\lambda$ reflects how the objective function changes as the constraint is varied.

---

Would you like more details on this method or other examples? Here's a quick list of related topics:

1. How to handle multiple constraints with Lagrange multipliers.
2. How to visualize the method geometrically.
3. Conditions for a local extremum in constrained optimization.
4. The role of the Lagrange multiplier in economics and physics.
5. Applying Lagrange multipliers in higher dimensions.

Tip: If you have multiple constraints, you can extend Lagrange multipliers by introducing a multiplier for each constraint and solving the resulting system.