4 to find the dimensions of the rectangle that have the maximum area. 2y A =?? f(x, y) = (2x)(2y) = 4xy

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1 Optimization Constrained optimization and Lagrange multipliers Constrained optimization is what it sounds like - the problem of finding a maximum or minimum value (optimization), subject to some other restrictions or constraints. Example: Suppose we have a rectangle inscribed within a given ellipse, say x2 to find the dimensions of the rectangle that have the maximum area. = 1. We would like A =?? 2y A =?? 2y 2x 2x The function that we wish to maximize, called the objective function, is the area function: f(x, y) = (2x)(2y) = xy The constraint is that x and y are related through the ellipse: We let g(x, y) be the constraint function = 1 1 = 0 g(x, y) = x2 1 Now, this particular example can be solved through single variable Calculus, and I ll suggest that you try that as a refresher. It ll also let us check the solution later on after we use the method of Lagrange multipliers. Solve x2 = 1 for y (use the positive square root). Substitute that into the area function, creating a function of one variable, x. Use the usual single variable Calculus process for finding a max or min value - differentiate, set equal to 0 and solve for x, and verify/test that you in fact have a maximum.

2 The solution is posted as a separate file, and linked in the lecture - you should get x = 3 2 [and can easily obtain y from there]. Now, back to Lagrange multipliers... If we look at the area as a function of two variables, we have a 3D surface. And just looking at f(x, y) = xy, we d see that that function by itself doesn t have a maximum value - as x and y get larger, the area grows without bound - so the constraint is needed to make sense of the problem. If we consider all possible values that the area could take on, e.g. an area of 1, 1.5, 2, , and so on, so f(x, y) = xy = 1 f(x, y) = xy = 1.5 f(x, y) = xy = 2 f(x, y) = xy = etc., etc., then we are looking at things in the form f(x, y) = xy = k i.e., the level curves of the function. Here is a contour plot showing some of the level curves of f(x, y): If we superimpose the constraint ellipse on the level curves, we see first of all that only some of the level curves intersect with the constraint, and therefore contain allowable values for x and y, and we can discard any that don t:

3 Now, it would help if I noted the heights that those level curves are at, which are the prospective values for area, because you d see that they are increasing as you move out from the center. So what we re really looking for is the level curve that just barely intersects the constraint equation - that s the one where the area will be maximal. Just barely intersects would be the same as tangent to. k = 12 k = 8 k = 5 k = 3 k = 1.5 k =.5 It looks like the maximal area that still intersects the constraint is when k = 12, so when xy = 12. Set xy = 12, so y = 3 x and sub into x2 = 1 : 9 + (3/x)2 = = 1 x + 81 = 36 x = 0 (2 9)(2 9) = = 0 x = 3 2

4 Now, we need to formalize that approach... Method of Lagrange multipliers The trick here is to use the gradient vectors, and recall the useful result that the gradient vector at any point is always orthogonal to the level curve at that point. So it s orthogonal to a vector tangent to the level curve in the plane of the curve. If you want the curves to be tangent to each other, they need to have the parallel tangent vectors...which means they need to have parallel gradient vectors. And parallel for vectors means the same vector, or scalar multiples of each other. So we need to find (x 0, y 0 ) that solve f(x, y) = λ g(x, y) and also satisfy the constraint equation g(x, y) = 0. Lagrange s Theorem Proof: Let f and g have continuous first partial derivatives such that f has an extremum at a point (x 0, y 0 ) on the smooth constraint curve g(x, y) = 0. If g(x 0, y 0 ) 0, then there is a real number λ such that f(x 0, y 0 ) = λ g(x 0, y 0 ) Let r(t) be a parameterization of the smooth curve given by g(x, y) = 0; r(t) = x(t)i + y(t)j with r (t) 0, where x and y are continuous functions of t on an open interval I. [Note that you don t have to find the parameterization to work the problem, but asserting that one exists is needed for the proof. The smooth part is where r (t) 0 comes in.] Define a function h(t) = f(x(t), y(t)). Since (x 0, y 0 ) is an extreme value of f, we know that h(t 0 ) = f(x(t 0 ), y(t 0 )) = f(x 0, y 0 ) must be an extreme value of h. So h (t 0 ) = 0. [h(t) is a single variable function, and single variable Calculus tells us its derivative is zero wherever it has a max or min value.] Since h(t) is a composite function, h(t) = f(x(t), y(t)), we apply the chain rule to get or dh dt = f dx x dt + f dy y dt h (t) = f x (x(t), y(t))x (t) + f y (x(t), y(t))y (t) = f(x(t), y(t)) r (t) Since h (t 0 ) = 0, f(x 0, y 0 ) r (t 0 ) = 0, and f(x 0, y 0 ) is orthogonal to r (t 0 ).

5 Also, we have already established that the gradient is orthogonal to the level curve of a function at a point: g(x 0, y 0 ) is orthogonal to the level curve of g passing through that point. Since r(t) is the curve g(x, y), r (t) must be tangential to that curve, and so g(x 0, y 0 ) is also orthogonal to r (t 0 ). Therefore, f(x 0, y 0 ) and g(x 0, y 0 ) must be parallel to (i.e. scalar multiples of) each other, and so there exists some λ such that Back to the example... f(x 0, y 0 ) = λ g(x 0, y 0 ) Use the method of Lagrange multipliers to maximize f(x, y) = xy subject to the constraint 1 = 0 Step 1: Find f for the objective function f(x, y) = xy, and g for the constraint function g(x, y) = x2 1 Step 2: Set up the equation f = λ g, and split that into two equations { fx = λg x f y = λg y

6 Step 3: Solve the system of three equations and three unknowns - the two equations that come from f = λ g, and also the constraint g(x, y) = 0. Occasionally you may be lucky enough to have a system of linear equations, but frequently, these will be nonlinear, and you need to resort to creative algebra.

7 In general, how do we know that the solution is the extremum we want? If the problem calls for a max, and we get one solution, what s telling us that that solution isn t in fact a min? Recall that with previous optimization techniques, there s been a test at the end to determine whether you have a max or min (recently, we had a second derivative type test for two variable optimization). The method guarantees an extremum of some sort - so to test that your solution (x 0, y 0 ) produces a maximum f(x 0, y 0 ), verifying that f(x 0, y 0 ) f(x, y) for all other (x, y) that satisfy the constraint reduces to the problem of verifying f(x 0, y 0 ) f(x, y) for just any one (x, y). Pick any other point on the graph of the constraint equation, and plug it in to f: Since (x, y) = (0, 2) is a point on x2 = 1, compute f(0, 2) = (0)(2) = 0. Since f( 3, 2) f(0, 2), f( 3, 2) = 12 is a maximum. 2 2 Finally... The process is more streamlined than it looks at first read - keep in mind we solved the same problem 3 ways and derived the method in the process. Implementing it isn t all that tedious. Look for a live example. The method extends to three (and more) variables - in the three variable case, the level curves become level surfaces. You ll see a couple of these in the suggested problems.

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