Lecture 6

The method works in a very similar way for more

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: blem will ‘constrain’ the variables in the right way, and we have for ourselves a solution to the problem we’re really interested in (the constrained problem). It is somewhat interesting to note that Lagrange’s approach is one of method – it is almost entirely based on off‐the‐shelf mathematical concepts. We’ll motivate this section by considering a simple constrained optimization problem with one constraint only. The method ‘works’ in a very similar way for more complicated problems, as long as the number of constraints is less than the number of critical points we are trying to find. Example: Let ∶ → . We want to choose , ∈ to maximize , (or minimize, depending on the context of the problem) subject to the constraint , 0. The general (unconstrained) problem would not be too difficult, if we had the proper assumptions on the function ∙ . But with this new angle, i.e., the restriction that , 0, we have a somewhat different problem in force. Side note #3: if is a linear function, as would be a budget constraint, the problem then can be written as a fairly easy problem of reduction the choice problem to one variable, say, . That is, first solve for in terms of using the condition , 0, and then find the critical point ∗ and finally, with ∗ we obtain ∗ . However, this approach doesn’t really help us much if a) is not linear and/or, b) there are more than 2 variables. Method of Lagrange, Cont. We form what is referred to as a Lagrangian: , , ≡ , , . The variable is an ancillary one, that is, a variable we have introduced to convert our constrained optimization problem into an unconstrained one. It is of very little general interest to us or to the problem at hand. We refer to it as a Lagrange Multiplier. The critical points , ∗ , ∗ of this problem satisfy: ∗, ∗ 2. ∗ ∗ ∗, ∗ ∗, ∗ 1. 3. ∗ ∗ ∗, ∗ , ∗ 0 0 0 These three first order conditions give us 3 equations in 3 unknowns, ∗ , ∗ , and ∗ . Notice that what we are finding are critical points ∗ , ∗ that, as we can see...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online