This note will de­rive the La­grangian mul­ti­pli­ers for an ex­am­ple prob­lem. Only cal­cu­lus will be used. The ex­am­ple prob­lem will be to find a sta­tion­ary point of a func­tion of four vari­ables if there are two con­straints. Dif­fer­ent num­bers of vari­ables and con­straints would work out in sim­i­lar ways as this ex­am­ple.

The four vari­ables that ex­am­ple func­tion de­pends on will be de­noted by , , , and . The two con­straints will be taken to be equa­tions of the form  0 and  0, for suit­able func­tions and . Con­straints can al­ways be brought in such a form by tak­ing every­thing in the con­straint’s equa­tion to the left-hand side of the equals sign.

So the ex­am­ple prob­lem is: Sta­tion­ar­ize means to find lo­ca­tions where the func­tion has a min­i­mum or a max­i­mum, or any other point where it does not change un­der small changes of the vari­ables as long as these sat­isfy the con­straints.

The first thing to note is that rather than con­sid­er­ing to be a func­tion of , you can con­sider it in­stead to be to be a func­tion of and and only two ad­di­tional vari­ables from , say and : The rea­son you can do that is that you should in prin­ci­ple be able to re­con­struct the two miss­ing vari­ables and given , , , and .

As a re­sult, any small change in the func­tion , re­gard­less of con­straints, can be writ­ten us­ing the ex­pres­sion for a to­tal dif­fer­en­tial as: At the de­sired sta­tion­ary point, ac­cept­able changes in vari­ables are those that keep and con­stant at zero; they have  0 and  0. So for to be sta­tion­ary un­der all ac­cept­able changes of vari­ables, you must have that the fi­nal two terms are zero for any changes in vari­ables. This means that the par­tial de­riv­a­tives in the fi­nal two terms must be zero since the changes and can be ar­bi­trary.

For changes in vari­ables that do go out of bounds, the change in will not be zero; that change will be given by the first two terms in the right-hand side. So, the er­ro­neous changes in due to go­ing out of bounds are these first two terms, and if we sub­tract them, we get zero net change for any ar­bi­trary change in vari­ables: In other words, if we pe­nal­ize the change in for go­ing out of bounds by amounts and at the rate above, any change in vari­ables will pro­duce a pe­nal­ized change of zero, whether it stays within bounds or not.

The two de­riv­a­tives at the sta­tion­ary point in the ex­pres­sion above are the La­grangian mul­ti­pli­ers or penalty fac­tors, call them     and     . In those terms for what­ever is the change in the vari­ables , and that means for what­ever is the change in the orig­i­nal vari­ables . There­fore, the change in the pe­nal­ized func­tion is zero what­ever is the change in the vari­ables .

In prac­ti­cal ap­pli­ca­tion, ex­plic­itly com­put­ing the La­grangian mul­ti­pli­ers and as the de­riv­a­tives of func­tion is not needed. You get four equa­tions by putting the de­riv­a­tives of the pe­nal­ized with re­spect to through equal to zero, and the two con­straints pro­vide two more equa­tions. Six equa­tions is enough to find the six un­knowns through , and .