MA 261 On LAGRANGE MULTIPLIERS based on Stewart’s textWe assume that you are familiar with the standard tests for extrema when wehave a function f of more than one independent variable; we usually study f(x, y)but the same ideas work for f (x, y, z), etc. The important thing is being clearon the difference between indepe ndent and non-independent variables. In thisintroduction we and examples we write the variable as (x, y)or(x, y, z), but usevector notation in describing the general principle.1 Independent VariablesSuppose we want to find the extrema of f(x, y)where(x, y) ranges over somedomain D (for convenience we always assume that f has continuous partials). Thestandard method is (a) to identify the possible interior extrema by finding points(x, y)where∇f(x, y)=0. This is where the issue of the variables being independentarises, since when they are independent, we may leave (x, y) in any direction andcheck whether f increases or decreases in that direction (that is why directionalderivatives are introduced.) we introduced directional derivatives in the previouslecture). Only if ∇f(x, y)=0 will the directional derivative of f in any directionfrom (x, y) be zero, which means that there is no obvious direction in which to gofrom (x, y)wheref would increase or decrease, and so (x, y) has to be considereda point at which f might attain an extremum.Step (a) is followed by (b): identifying the possible extrema taken on the bound-ary of D, the situation discussed here.2 Dependent variablesThe situation is different when we look for points on the boundary of D at whichthe extrema occur; this is analogous to checking at the ‘endpoints’ when f is definedon [a, b], and f might have an extremum at either x = a, x = b without f0(a)orf0(b) vanishing. In two variables this usually that means that a given portion of ∂D(the boundary of D) is represented by one of the forms (i) y = g(x), (ii) x = g(y),or (iii)(x, y)=(x(t),y(t)), so that t is a parameter.It is important to see in that situation we have lost an independent variable!In case (i) we are saying that only x is independent, so that once x is given weknow y – that can’t occur were the variables inependent. In (ii) y is the onlyindependent variable, and in (iii) only t is independent. So, for example, in (i),we have f (x, y)=f(x, g(x)) = F (x), in case (ii) f (x, y)=F (y), and in (iii)f(x, y)=f(x(t),y(t)) = F (t) [I am using the letter F for a different function ineach case]. When (x, y) ∈ ∂D, the variables are not independent, so we are notable to move (x, y) in any direction we choose and remain on the boundary. Thatis the reason f might have an extremum at (x, y) without ∇f(x, y) being zero.(When we consider a function f of n variables defined in a region D, there willin general be n −1 independent variables on ∂D.)In our earlier homework, when doing these problems without Lagrange multi-pliers, we would simply study f on this portion of ∂D by going back to one-variablemaximum/minimum techniques from calculus, and so identify all possible extremaon the boundary. This can lead to many cases, when the boundary has pieces whichare of various types (i)–(iii). [There will be examples given.]3Na¨ıve way to handle boundary situationThe Lagrange method is a more insightful – and simple – way of handling thesituation that the number of independent variables is reduced — that is what wemean by there being a constraint. As we have mentioned, the constraint is thateither y = g(x),x = h(y)orx = x(t),y = y(t), etc., which as we have observedreduces the number of independent variables.4 Lagrange’s insightLagrange had an insight that uses geometry and some elementary vector analysis.The exposition I am giving here covers §14.8 in (my opinion) a simpler way, andalways leads to “one less equation” in a system of simultaneous equations. Nomatter how we approach these problems, solving these systems requires methodsimprovised for each problem – we do not use any general method.Our problem is toextremize f(x) subject to the ‘constraint’ g(x)=0:here x =(x1,x2,...,xk) is a vector.It is governed by an elementary fact about vectors.Principle: Let x =(x1,x2...,xk) and y =(y1,y2,...,yk) be be vectors (in ourcontext usually k =2or 3.) Then x k y if and only if all ratios xi/yi(i =1, 2,...,k)are the same; the understanding is that if we have xj=0for some j, then thecorresponding entry yjmust also be zero.(You should check with a few examples: for what λ is (1, 4, 9) k (−2, −8,λ)? Howwould your answer change if you replace only 4 with 6? Or 9 with 0?)Let us apply this to the situation of extremizing f subject to our constraintg =0, and let x =(a, b) be a point we are testing as a possible extrema. ThenLagrange: in order that f have an extrema at x it is necessary that∇f(x) k∇g(x). (1)Note. The word multiplier arises from the usual formulation of this principle: itasserts that at a potential extrema we have the equation∇f(x)=λ∇g(x),where λ is a scalar. You should check that these two ways of expressing the Lagrangeprinciple are the same; the (slight) advantage in the formulation I prefer is that thescalar λ usually has no physical significance, and in practice (that is, homeworkproblems!) means introducing λ simply gives is an extra equation to consider, withno extra information.Once we identify the points where (1) holds, we have to compute f at each, andthen the maximum of f among points which satisfy the constraint will be wheref takes on the largest value among these points; a similar remark will lead to theminimum of f.5 Some examplesLet’s see how this works with some of the homework and examples in the text.We should always identify f and g. We then use the Lagrange principle, and whatwill come out is that the coordinates at a possible extremum must satisfy a certainrelation. In other to find the exact values of the coordinates at these possibleextrema, however, we have to return to the constraint g = 0, since if our constraintwere g = c, c a constant (with c 6= 0), the Lagrange equation would be the same,and so would be a relation between the variables. (Our first example is an exceptionto this principle, because the relation (1) is more subtle, since the coordinates atthe extrema will be zero.) I start with a complicated example, which was doneearlier by other methods.P. 962, No. 39. Find a point on the surface z2= xy + 1 closest to the origin.Step I: Find f, g. Well, it is f that we