Review: critical point or equivalently f a,

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1 Review: a b f f a b f a b critical point or equivalentl f a, b A point, is called a of if,, 0 A local ma or local min must be a critical point (but not conversel) 0 D iscriminant (or Hessian) f f D f f f f f D a, b 0 and f ( a, b) 0 (or f ( a, b) 0) ( a, b) is a local min D a, b 0 and f ( a, b) 0 (or f ( a, b) 0) ( a, b) is a local ma D a, b 0 ( a, b) is a saddle point D a, b 0 inconclusive Etreme Value Theorem f, is continuous in some f assumes its absolute maimum and closed bounded region in minimum at some point in To find the absolute ma and min a) Find critical points inside S S R S b) On each piece of the boundar of S find the critical points (for the function restricted to the bounda) c) Evaluate function at points in a) and b), and at "corner points" The largest and smallest of these values are the absolute ma and min.

2 Problem : The volume is fied, so z 1 The cost is C(,, z) 10z 5z 10 10z 15z 10 10z We can substitute z : C Minimize C 10 subject to 0, C or C A bo is to be constructed with 5 wooden sides and a glass front. The wood costs $5 per square foot and the glass $10 per square foot. The bo should have a volume of 1 cubic feet. What dimensions minimize the cost. Let and be the width and depth of the base and z the height of the bo or critical point (, ) (,3) 1 1 height z 6 hence width and height and depth or 8, hence 18

3 Wh does the second derivative test work? Then f, 0 and f, 0 Assume the critical point is at (, ) f must have a local min along ever straight line through ( 0, 0). Let (, ) ( sa, sb) be a straight line in the direction of ( a, b) 0 0 Define g( s) f ( sa, sb). 0 0 The function g must have a local min at s 0 : g '( s) f ( sa, sb) a f ( sa, sb) b g f a f ab f ab f b ''(0) g '(0) f (, ) a f (, ) b g ''(0) f (0,0) a f (0,0) ab f (0,0) b 0 for all ab, a Divide b b and set = : f ( 0, 0) f ( 0, 0) f ( 0, 0) 0 b Now recall that a b c 0 has no real solution if ac b 0 If then also a 0, we have a b c 0 for all So if f ( 0, 0) 0, f ( 0, 0) f ( 0, 0) ( f ( 0, 0)) 0 then g ''(0) 0

4 14.8 Lagrange Multipliers

5 Goal: To maimize a function f, subject to the constraint g, k.

6 Goal: To maimize a function f, subject to the constraint g, k. Geometricall this means that we are looking for a point, 0 0 on the graph of the constraint curve at which f, is as large as possible. To help us locate such a point we can contruct a contour plot of f,. Each point of intersection of g, k with a level curve of f is a candidate for a solution, since these points lie on the constraint curve. The largest value of f occurs when these curves just touch each other, when the have a common tangent line. At this point,, the gradient of f 0 0 is parallel to the gradient of g. f, g, for some scalar

7 Another eplanation : Maimize the function f, subject to the constraint g, k. Claim: We need f, g, for some Let r( t) ( ( t), ( t)) be a parametrization of the curve g, k. Find ma of g( t) f ( ( t), ( t)) Need g'( t) 0 but g '( t) f ( ( t), ( t)) '( t) f ( ( t), ( t)) '( t) g '( t) f '( t), '( t) f r '( t) 0 So f is orthogonal to r ' But g is also orthogonal to the level curve g k So g is orthogonal to r ' as well Hence f is parallel to g So f g for some

8 A wa to simplif the computation: f, g, f g ,, ) write the constraint as g 0 (subtract the constant k) F f g set up the function,,,, 3 solve F 0 F 0, F 0, and F 0 4 find all such points that satisf these equations and evaluate The largest value of of f subject to the constraint g. The smallest value of f at these points of f subject to the constraint g. Maimize a function f, subject to the constraint g, k. f f will be the maimum will be the minimum equivalent to the constraint g 0

9 Eample: Find the maimum and minimum of the function f (, ) restricted to the ellipse 4 4 constraint: g(, ) Find critical points of F(,, ) f g ( 4 4) F 4 0 or ( ) 0 F 8 0 or (1 4 ) 0 1 a) 0 and from constraint: 1 4 b) and 0 from constraint: F ( 4 4) 0 or 4 4 critical points: (0,1),(0, 1),(,0),(,0) notice, the value of is not relevant! function values: f (0,1) 3, f (0, 1) 3, f (, 0) 10, f (, 0) 10 maimum is 10 and minimum is 3 PUZZLE: 4 4, hence f (, ) (4 4 ) hence f ' 14 0 onl if 0 and hence. Whats wrong here? or f (, ) 7 10

10 A picture: The maimum and minimum of the function f (, ) restricted to the ellipse 4 4 critical points: (0,1),(0, 1),(,0),(,0) function values: f (0,1) 3, f (0, 1) 3, f (, 0) 10, f (, 0) 10

11 Eample: Find the point on the circle 45 that is closest to 1,. Notice: minimizing f and minimizing f is the same minimize f 1 constraint: g 45,, 1 45 F F 1 0 F 0 F two critical points: 3, 6 and 3, f 3, the actual distance is 0 5 f 3, the actual distance is ,6 The point on the circle 45 that is closest to 1, is.