Lecture 15. Global extrema and Lagrange multipliers. Dan Nichols MATH 233, Spring 2018 University of Massachusetts

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1 Lecture 15 Global extrema and Lagrange multipliers Dan Nichols MATH 233, Spring 2018 University of Massachusetts March 22, 2018 (2) Global extrema of a multivariable function Definition Let f(x, y) be a continuous function of two variables on domain D. f(x, y) has an absolute maximum (or global maximum) at (a, b) if f(a, b) f(x, y) for all (x, y) in D. f(x, y) has an absolute minimum (or global minimum) at (a, b) if f(a, b) f(x, y) for all (x, y) in D. A absolute extremum (or global extremum) is a point that s either an absolute maximum or an absolute minimum. Theorem If an absolute extremum of f exists, it must occur either at a critical point in the interior of D or on the boundary of D.

2 (3) The EVT for multivariable functions Theorem (Extreme value theorem for multivariable functions) Suppose f(x, y) is continuous on a closed, bounded domain. Then f has both an absolute maximum and an absolute minimum somewhere in this domain. To find the absolute extrema of f(x, y) on a closed, bounded domain D, we follow these steps: 1. Find the critical points (x, y) = (a, b) in the interior of D 2. Make a list of the values of f(a, b) for all critical points (a, b) 3. Find the max. and min. values of f on the boundary of D (a) Break up curve into nice pieces, turn f into a single-variable function on each piece (b) Use single-variable local extrema algorithm to find extrema on each piece (remember endpoints). Add them to the list 4. Highest value on the list is the absolute maximum. Lowest value on the list is the absolute minimum. (4) Global extrema of a multivariable function: example Example 1: Find the global extrema of f on the domain D, where f(x, y) = x 6 + y 6 2xy + 6 D = {(x, y): 1 x 1, 1 y 1} y x Need to check Critical points in interior Four line segments (boundary) Four corners (endpoints of boundary curves) The solution to this problem is in a separate document on the section website.

(5) Lagrange multipliers Notice that one step in our global extrema algorithm is to find the maximum/minimum value of f(x, y) over a curve in the xy-plane. We have two functions, f(x, y) and g(x, y).

3 (5) Lagrange multipliers Notice that one step in our global extrema algorithm is to find the maximum/minimum value of f(x, y) over a curve in the xy-plane. We have two functions, f(x, y) and g(x, y). We want to find a point (x, y) which makes the value of f(x, y) as big/small as possible while making sure that g(x, y) = k. We call this condition the constraint. Think of g(x, y) = k as an implicit curve in the xy-plane, and z = f(x, y) as a surface above/below it. Find points where the surface is highest/lowest without leaving the curve. (6) Lagrange multipliers Up until now, we ve done this by parameterizing (each piece of) the curve g(x, y) = k with functions x(t), y(t). Value of f(x, y) (on the curve) becomes a function of t: f(t) = f(x(t), y(t)) Use single-variable calculus to maximize or minimize this function. y x Instead, let s take advantage of partial derivatives and the gradient.

4 (7) Lagrange multipliers Here s a curve g(x, y) = k together with a contour map of another function f(x, y). Suppose z = m is the maximum value reached by f on the constraint curve g(x, y) = k. Then f(x, y) = m is the highest level curve of f that intersects the constraint curve g(x, y) = k. At the point where these two curves intersect, they are tangent to each other (same tangent line). (8) Lagrange multipliers level curves of f(x, y) (assume height increases as you move towards center) y curve g(x, y) = 0 possible max./min. of f on blue curve tangent lines gradients f x Where g is tangent to a level curve of f, the gradients are parallel.

5 (9) Lagrange multipliers So if (a, b) is a point on the constraint curve g(x, y) = k where f(x, y) is maximized or minimized then the constraint curve g(x, y) = k and the level curve of f(x, y) through (a, b) have the same tangent line at (a, b). Therefore the gradients f(a, b) and g(a, b) are parallel (because the gradient is always orthogonal to the level curve) Theorem If f(x, y) has a maximum or minimum value at (x, y) = (a, b) subject to the constraint g(x, y) = k, then f(a, b) = λ g(a, b) for some scalar λ 0. We call the number λ a Lagrange multiplier. We always assume g(x, y) = k is a simple curve, i.e. g(x, y) 0 on the curve. Otherwise this might not work. (10) Lagrange multipliers Here s how to use the Lagrange multiplier method to find extreme values of f(x, y) subject to the constraint g(x, y) = k: 1. Compute f(x, y) and g(x, y). 2. Write down three equations in the variables x, y, and λ: one from constraint, two from components of the equation f(x, y) = λ g(x, y). g(x, y) = k f x (x, y) = λg x (x, y) f y (x, y) = λg y (x, y) 3. Find all triples (x, y, λ) which satisfy all 3 equations at once. 4. Evaluate f at all the points (x, y) you found. The highest value you find is the maximum and the lowest value is the minimum.

6 (11) Lagrange multipliers: example Example 2: Find the maximum and minimum values of f(x, y) = y 2 x 2 subject to the constraint x2 4 + y2 = 1. f(x, y) = y 2 x 2 x y2 = 1 y 4 z 2 0 x x y 2 (12) Lagrange multipliers: example Example 2: (cont.) Find the maximum and minimum values of f(x, y) = y 2 x 2 subject to the constraint x2 4 + y2 = 1. (Step 1) compute gradients f(x, y) = 2x, 2y g(x, y) = 1 x, 2y 2 (Step 2) write down 3 simultaneous equations: the constraint g(x, y) = k, and the two components of f(x, y) = λ g(x, y). x y2 = 1 2x = λ 1 x 2y = λ2y 2

7 (13) Lagrange multipliers: example Example 2: (cont.) (Step 3) solve the system of equations: x y2 = 1 2x = λ x 2y = λ2y 2 In general: Pick one equation and solve for one variable in terms of the other two variables. Then plug that expression into the other equations. Helpful tip 1: Break things down into cases. Say either this variable must be [something], or else that variable must be [something]. Helpful tip 2: View each equation as a curve. Visualize the curves and figure out where all 3 intersect. (14) Lagrange multipliers: example Example 2: (cont.) (Step 3) solve the system of equations: x y2 = 1 2x = λ x 2y = 2λy 2 To satisfy 2y = 2λy, we need either y = 0 or λ = 1. To satisfy 2x = λ 2 x, we need either x = 0 or λ = 4. Once we know either x or y, the equation x 2 /4 + y 2 = 1 determines the other one of those variables. Ways to satisfy the equations: λ = 1, λ = 4 Impossible y = 0, x = 0 Doesn t satisfy constraint y = 0, λ = 4, x = ±2 λ = 1, x = 0, y = ±1 So the maximum and minimum values of f on the constraint must be somewhere among the points ( 2, 0), (2, 0), (0, 1), (0, 1).

8 (15) Lagrange multipliers: example Example 2: (cont.) f(x, y) = y 2 x 2, constraint x2 4 + y2 = 1. Possible max./min. values at ( 2, 0), (2, 0), (0, 1), (0, 1) (Step 3) compute the value of f at each of these points, and compare them. The maximum value of f subject to this constraint is 1, which occurs at (0, 1) and (0, 1). The minimum value of f subject to this constraint is -4, which occurs at ( 2, 0) and (2, 0). (a, b) f(a, b) ( 2, 0) -4 (2, 0) -4 (0, 1) 1 (0, 1) 1 (16) Lagrange multipliers: example Another way to solve the equations: use λ as a parameter. 1. Solve the λ equations to get x,y as functions of λ. 2. Plug these functions for x and y into the constraint, solve for λ. 3. Take the values of λ you found, plug them back into the λ equations to get points (x, y) which satisfy all 3. Yet another way: solve both of the λ equations for λ and set the results equal to each other to get a new equation. Now you re solving a system of only two equations (the new equation and the constraint) in only two variables (x and y). Unfortunately there s no perfect algorithm that works every time... you have to experiment and be creative.

9 (17) Lagrange multipliers: 3D: example The Lagrange multipliers technique also works for functions of more than two variables. Assume that maximum and minimum values of f(x, y, z) subject to the constraint g(x, y, z) = k exist Think of g(x, y, z) as an implicitly-defined surface and f(x, y, z) as a function on 3D space. Also assume g 0 on the surface g(x, y, z) = k To find these values, 1. Find all values of x, y, z, λ such that g(x, y, z) = k and f(x, y, z) = λ g(x, y, z). (Four scalar equations) 2. Evaluate f at all the points (x, y, z) you found. The largest of these values is the maximum and the smallest is the minimum. (18) Lagrange multipliers: 3D: example Example 3: Find the maximum and minimum values of f(x, y, z) = xyz subject to the constraint x 2 + 2y 2 + 3z 2 = 6.

10 (19) Lagrange multipliers: 3D: example Example 3: (cont.) Find the maximum and minimum values of f(x, y, z) = xyz subject to the constraint x 2 + 2y 2 + 3z 2 = 6. The solution to this problem is in a separate document on the section website. (20) Lagrange multipliers and global extrema Example 4: Find the absolute maximum value of f(x, y) = x 2 + y 2 6x + 6y on the domain x 2 + y Remember: to find global extrema of f(x, y) on a closed domain D, we need to check 1. critical points in the interior of D 2. the boundary of D If the boundary of D is a nice implicit curve, we can use Lagrange multipliers to look for boundary extrema. y x

11 (21) Lagrange multipliers and global extrema Example 4: (cont.) Find the absolute maximum value of f(x, y) = x 2 + y 2 6x + 6y on the domain x 2 + y f(x, y) = 2x 6, 2y + 6 = 0 The only critical point is (3, 3) which lies in the interior of D. Constraint: g(x, y) = x 2 + y 2 = 36, so g(x, y) = 2x, 2y Lagrange multiplier equations: 2xλ = 2x 6 2yλ = 2y + 6 x 2 + y 2 = 32 From the first two equations we get λ = 1 3 x = y, so we need y = x Plug this into the constraint and we find x = ±4. Remember y = x so the possible boundary extrema are at ( 4, 4) and (4, 4). Compare all possible extrema: absolute max. is f( 4, 4) = 80. (x, y) (3, 3) ( 4, 4) (4, 4) f(x, y) (22) Lagrange multipliers and global extrema There s actually a dumb shortcut we can use to easily find boundary extrema in the previous example (and other problems like it). Example 4: (cont.) Find the absolute maximum value of f(x, y) = x 2 + y 2 6x + 6y on the domain x 2 + y On the boundary we can rewrite the function as f(x, y) = 32 6x + 6y because we know that (on the boundary) x 2 + y 2 = 32. This makes things much easier. You don t even really need to use Lagrange multipliers. Use this when f(x, y) and g(x, y) are similar, i.e. when you can easily simplify f(x, y) using the constraint equation. You still need to handle the interior of D (critical points) the regular way though.

12 (23) Lagrange multipliers and global extrema Example 5: Let D be the region defined by x 2 + (y 1) 2 4. Find the global extreme values of f(x, y) = x 3 + xy 2 x on D. f(x, y) = 3x 2 + y 2 1, 2xy Critical points: need 2xy = 0, so either x = 0 or y = 0 Plug x = 0 into the other equation: y 2 1 = 0 means y = ±1 Plug y = 0 into the other equation: 3x 2 1 = 0 means x = ± 1/3 So the critical points are (0, 1), (0, 1), ( 1/3, 0 ), and ( 1/3, 0 ). Notice that (0, 1) is not in the interior of D, it s on the boundary. That just means we ll see it again when we check for boundary extrema, which is the next step. (24) Lagrange multipliers and global extrema Example 5: (cont.) Global extrema of f(x, y) = x 3 + xy 2 x on x 2 + (y 1) 2 4 Remember f(x, y) = 3x 2 + y 2 1, 2xy The constraint is g(x, y) = x 2 + (y 1) 2 = 4, so g(x, y) = 2x, 2y 2 Lagrange multiplier equations: 2xλ = 3x 2 + y 2 1 (2y 2)λ = 2xy x 2 + (y 1) 2 = 4 Solve each of the first two equations for λ, set them equal: 3x 2 + y 2 1 2x = 2xy 2y 2 Cross-multiply: 2(3x 2 y 3x 2 + y 3 y 2 y + 1) = 4x 2 y Simplify and write as an expanded polynomial in x: (y 3)x 2 + (y 3 y 2 y + 1) = 0 (y 3)x 2 + (y + 1)(y 1) 2 = y x = ±(y 1) 3 y. Now we just need to find points (x, y) which satisfy both this and the constraint.

13 (25) Lagrange multipliers and global extrema Example 5: (cont.) Global extrema of f(x, y) = x 3 + xy 2 x on x 2 + (y 1) 2 4 Boundary extrema must satisfy both x = ±(y 1) 1+y 3 y and the constraint. Plug that equation into the constraint: [ ] y ±(y 1) + (y 1) 2 = 4 3 y (y 1) y 3 y + (y 1)2 = 4 (y 1) 2 ( 1 + y 3 y + 1 ) = 4 (y 1) y = 4 (y 1) 2 = 3 y y 2 y 2 = 0. Factor: solutions are y = 1, y = 2. Plug these back into the equation 1+y x = ±(y 1) 3 y get (0, 1) (already found) and ( ± 3, 2 ) (x, y) f(x, y) (0, 1) 0 (0, 1) 0 ( ) 1/3, 0 ( 1/3, ) 0 ( ) ( 3, 2 ) 3, min 6 3 max (26) Lagrange multipliers: multiple constraints Sometimes there may be more than one constraint. Suppose you want to maximize f(x, y, z) subject to g(x, y, z) = 0 and h(x, y, z) = 0. Need to check all points (x, y, z) which satisfy both constraints and which satisfy the equation g(x, y, z) = 0, h(x, y, z) = 0 f(x, y, z) = λ g(x, y, z) + µ h(x, y, z) For instance, with three dimensions and two constraints you need to solve a system of 5 scalar equations in 5 variables. (You don t have to worry about this in MATH-233.)

14 (27) Summary EVT: if f is continuous on a closed bounded domain D, then it has an absolute max. and min. on D Find the global extrema of f on D by checking extrema on the boundary of D and critical points in the interior of D Lagrange multiplier method can be used to maximize a multivariable function subject to some constraint. This is useful for checking the boundary when you re looking for global extrema Hard part is, again, solving systems of nonlinear equations (28) Homework Paper homeworks #14 AND #15 due Tuesday WebAssign homeworks 14.7, 14.8 due Wednesday, 11:59 PM Midterm 2: Thursday 4/5, 7-9 PM, location TBD Covers sections , Practice problems will be posted on the website soon

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

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