PX275 - A5 - lagrange multipliers

• typically, given a function, $f(x,y)$, a maxima or a minima can be found via the partial derivatives, ie:

• or, it can be found in terms of the total derivative, ie:

• if the maximum or minimum of the function, $f(x,y)$, with respect to a constraint, $y=1-x$, is to be found, the constraint needs to be rewritten as a function, $g(x,y)=x+y=1$

• note: the constraint being exactly equal to a constant is critical for this method

• at the max of $f(x,y)$ subject to the constraint, $\frac{\partial f}{\partial x}\ne 0;\frac{\partial f}{\partial y}\ne 0$, but because $g(x,y)=constant$, it can be said that $dg=\frac{\partial g}{\partial x}dx+\frac{\partial g}{\partial y}dy=0$

• without a formal proof, a statement for the minimum/maximum condition is:

where, $\lambda$ is an undetermined constant, called the lagrange multiplier

• the total differential will be:

• this system of simultaneous equations can be solved to find $x,y\text{and }\lambda$