Statistical Thermodynamics and Rate Theories/Lagrange multipliers

Constrained Optimization
When constraints are imposed on a system, Lagrange multipliers can be incorporated to determine the maximum of the multivariable function. The new process for determining maximum of this function within the constraint becomes:


 * 1) Write the constraint as a function. i.e., $$g(x,y) = c$$
 * 2) Define a new equation. $$f(x,y) + {\alpha} ({g(x,y)} - {c})$$ where $$\alpha$$ is an undefined constant
 * 3) Solve this set of equations to find the maximum, using the previous three steps for determining the maximum of an unconstrained system.

For example, suppose we have a function $$f(x,y) = -{2x}^2 - {y}^2 - {xy} - {10y}$$ and we impose the following constraint upon the function: $$y = x + 5$$

The constraint would be written as $$g(x,y) = x + y = 5$$

We would then define the new equation following the constraint as $$(-{2x}^2 - {y}^2 - {xy} - {10y}) + {\alpha} ({x} + {y} - {5})$$

Next, we take the partial derivative with respect for both x and y, set it to zero, and solve for x and y.

First of all taking the partial derivative with respect for x set to zero:

$$\frac{\partial}{\partial x} [(-{2x}^2 - {y}^2 - {xy} - {10y}) + {\alpha} ({x} + {y} - {5})] =0$$

Evaluating this will give:

$${\alpha}-4x-y =0$$

$${\alpha} =4x+y$$

Now, the partial derivative with respect for y set to zero will be taken:

$$\frac{\partial}{\partial y} [(-{2x}^2 - {y}^2 - {xy} - {10y}) + {\alpha} ({x} + {y} - {5})] =0$$

$${\alpha}-x-2y-10 =0$$

$${\alpha}= x+2y+10$$

This leads to the following system of equations that can be solved to determine the maximum:

$${\alpha} =4x+y$$

$${\alpha}= x+2y+10$$

$$x+y =5$$

From Solving this set of equations the maximum subjected to the constraint of $$x+y =5$$ is found to be:

$$(x,y) = \left(\frac{-35}{8} ,\frac{5}{8}\right)$$

Unconstrained Optimization
Determining the maximum of an unconstrained system follows very similar steps it just will not have a lagrange multiplier as the system is not subjected to the

maximum along a given line, rather the maximum of the system itself. The steps to solve an unconstrained system becomes:


 * 1) Calculate the partial derivatives
 * 2) set them to zero
 * 3) solve for the variables

For example given the same function of $$f(x,y) = -{2x}^2 - {y}^2 - {xy} - {10y}$$ first the partial derivatives will be calculated to be:

$$(\frac{\partial f}{\partial x})_y =-4x-y$$

$$(\frac{\partial f}{\partial y})_x =-2y-x$$

Let both partial derivatives be zero:

$$-4x-y =0$$

$$-2y-x =0$$

Finally the variables will be solved for. Giving a maximum that is different than that of the constrained system:

$$(x,y) =\left(\frac{10}{7} ,\frac{-40}{7}\right)$$