1. Show that the rectangle of maximum area that has a given perimeter p is a square.

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1 Constrained Optimization - Examples - 1 Unit #23 : Goals: Lagrange Multipliers To study constrained optimization; that is, the maximizing or minimizing of a function subject to a constraint (or side condition). Reading: Section In the previous section, we saw some of the difficulties of working with optimization when there are multiple variables. Many of those problems can be cast into an important class of problems called constrained optimization problems, which can be solved in an alternative way. Examples of Problems with Constraints 1. Show that the rectangle of maximum area that has a given perimeter p is a square. The function to be maximized: The constraint: A(x, y) = xy 2x + 2y = p Constrained Optimization - Examples Find the point on the sphere x 2 + y 2 + z 2 = 4 that is closest to the point (3, 3, 5). Here we want to minimize the distance subject to the constraint D(x, y, z) = (x 3) 2 + (y 3) 2 + (z 5) 2 x 2 + y 2 + z 2 = 4 Lagrange Multipliers Lagrange Multipliers - 1 To solve optimization problems when we have constraints on our choice of x and y, we can use the method of Lagrange multipliers. Suppose we want to maximize the function f(x, y) subject to the constraint g(x, y) = k. Consider the relative positions of their level curves:

2 Lagrange Multipliers - 2 Lagrange Multipliers - 3 On the above diagram, locate the maximum and minimum of f(x, y) (label with M and m respectively) on the level curve g(x, y) = k. Draw in the gradient vectors f and g at M and m. Describe, in words, the relationship between f(m) and g(m), and also between f(m) and g(m). The Lagrange Multiplier Method - 1 The Lagrange Multiplier Method - 2 Description of the Lagrange Multiplier Method Step 1. Find all values of (x, y) and λ such that f(x, y) = λ g(x, y) g(x, y) = k Step 2. Evaluate f at all the points (x, y) obtained in Step 1. (The largest value is the maximum and the smallest is the minimum.) Note that Step 1 really amounts to solving three equations in three unknowns (x, y, and λ). The equations can be found by rewriting the gradient equations as follows: f x (x, y) = λg x (x, y) f y (x, y) = λg y (x, y) g(x, y) = k The same method applies to functions of three (or more) variables. In this case of three variables, we would solve four equations in four unknowns.

3 Example: values of Lagrange Multiplier Method - Linear Constraint - 1 Consider the problem of finding the maximum and minimum f(x, y) = x 1y + x2 + y 2, 2 subject to the constraint x + y = 1. Sketch the meaning of the constraint. Lagrange Multiplier Method - Linear Constraint - 2 Write down the three equations obtained by the method of Lagrange multipliers. Lagrange Multiplier Method - Linear Constraint - 3 Lagrange Multiplier Method - Linear Constraint - 4 Solve these equations, and compare the values at the resulting points to find the maximum and minimum values

4 Lagrange Multiplier Method - Non-Linear Constraint - 1 Example: Find the points on the curve x 4 + y 4 = 1 that are closest to and furthest from the origin. Lagrange Multiplier Method - Non-Linear Constraint - 2 x 4 + y 4 = 1 Lagrange Multiplier Method - Non-Linear Constraint - 3 Lagrange Multiplier Method - Non-Linear Constraint - 4 x 4 + y 4 =

5 The Meaning of the Lagrange Multiplier The Meaning of the Lagrange Multiplier - 1 λ is called the Lagrange multiplier. It has its own significance, as can be seen from the following discussion. Recall that f(x, y) represents the rate of increase of the value of f if you move from (x, y) in the direction indicated by the gradient of f. Similarly, g(x, y) represents the rate of increase of the value of g if you move in the direction of the gradient of g. If f(x, y) = λ g(x, y) and λ >, then f(x, y) and g(x, y) have the same direction. (If λ < they have opposite directions.) If the vectors are parallel, then the value of λ represents the ratio λ = f(x, y) g(x, y) Thus, λ gives the rate of increase of f divided by the rate of increase of g. In other words, The Meaning of the Lagrange Multiplier - 2 λ gives the approximate increase in the optimum value of f when the value of constraint g is increased by 1. The Lagrange Multiplier - Labour and Capital - Part 1-1 Example (Labour and Capital) Suppose that the quantity q of a product depends on the number of workers, W, and the number of units of capital investment, K, and is represented by the Cobb-Douglas function q = 6W 3/4 K 1/4 In addition, labour costs are $1 per worker, capital costs are $2 per unit, and the budget is $3,2. We will ask several questions about this model. Use the method of Lagrange multipliers to find the optimum number of workers and optimum number of units of capital. The Lagrange Multiplier - Labour and Capital - Part 1-2 q = 6W 3/4 K 1/4

6 The Lagrange Multiplier - Labour and Capital - Part 1-3 The next part of the problem involves the marginal productivity of labour and the marginal productivity of capital. These concepts have to be translated into mathematical terms before the problem can be attempted. The marginal productivity of labour refers to the extra amount that would be produced if W were increased by one. That is, it is the value of q when W = 1. This means it is also the same as q W The Lagrange Multiplier - Labour and Capital - Part 1-4 Since a change of 1 is presumably very small compared to the value of W, we can use the fact that q W q, and interpret the marginal productivity of labour W to be the partial derivative q. Similarly, the marginal productivity of capital W should be interpreted as q K. Thus, marginal productivity of labour = q W marginal productivity of capital = q K The Lagrange Multiplier - Labour and Capital - Part 1-5 Check that at the optimum values of W and K, the ratio of the marginal productivity of labour to the marginal productivity of capital is the same as the ratio of the cost of a unit of labour to the cost of a unit of capital. The Lagrange Multiplier - Labour and Capital - Part 2-1 Recompute the optimum values of W and K when the budget is increased by $1. Check that increasing the budget by $1 allows the production of λ extra units of the good, where λ is the Lagrange multiplier.

7 4 The Lagrange Multiplier - Labour and Capital - Part 2-2 The Lagrange Multiplier - Labour and Capital - Part