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:

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

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:

$(x,y)=(\frac{10}{7},\frac{-40}{7})$ Template:BookCat