18 02A Topic 29 Second derivative test Lagrange multipliers Read TB 19 7 19 8 Review w 0 critical point Second derivative test for critical point x0 y0 Let wxx 0 A wxy 0 B wyy 0 C If AC B 2 0 then A 0 minimum A 0 maximum If AC B 2 0 then saddle If AC B 2 0 then test fails Example w x3 3xy y 3 wx 3x2 3y wy 3x 3y 2 w h3x2 3y 3x 3y 2 i Critical points 3x2 3y 0 y x2 Substitute this into 3x 3y 2 0 x4 x 0 x 0 1 critical points are 0 0 1 1 wxx 6x wxy 3 wyy 6y AC B 2 36xy 9 0 0 is a saddle and 1 1 is a minimum Reasoning Second order approximation at 0 0 w w 1 2w 2w 1 2w 2 w w0 x y x xy y2 x 0 y 0 2 x2 0 x y 0 2 y 2 0 More simply w wx 0 x wy 0 y 21 wxx 0 x2 wxy 0 xy 21 wyy 0 y 2 At a critical point this since wx 0 wxx A etc becomes w 21 Ax2 2Bxy Cy 2 Complete square w A x B y 2 A1 AC B 2 y 2 A So if AC B 2 0 and A 0 then w 0 minimum If AC B 2 0 and A 0 then w 0 maximum If AC B 2 0 then w varies saddle Examples Use these to remember the rules i z x2 y 2 min at 0 0 A 2 B 0 C 2 AC B 2 4 0 and A 0 ii z x2 y 2 max at 0 0 A 2 B 0 C 2 AC B 2 4 0 and A 0 iii z y 2 x2 saddle at 0 0 A 2 B 0 C 2 AC B 2 4 0 iv z xy saddle at 0 0 A 0 B 1 C 0 AC B 2 1 0 1 General example z ax2 2bxy cy 2 crit pt at 0 0 2 A a B b C c AC B 2 ac b2 0 continued 1 18 02A topic 29 2 Lagrange multipliers Problem Minimize w f x y z constrained by g x y z c Sphere example Minimize w y constrained to x2 y 2 z 2 1 Example Box No top sides double thick bottom triple thick volume 3 What s the smallest amount of cardboard you can use Dimensions x y z Cardboard w 4xz 4yz 3xy Constraint V xyz 3 Lagrange solution Critical point f g constraint g x y z c Sphere example f h0 1 0i g h2x 2y 2zi f g h0 1 0i h2x 2y 2zi x z 0 Constraint y 1 Gives min and max Box example f h4z 3y 4z 3x 4x 4yi V hyz xz xyi Lagrange h4z 3y 4z 3x 4x 4yi hyz xz xyi xyz 3 Solve symmetrically 4z 3y 4z 3x 4x 4y xyz 3 yz xz xy 4 y 4 y 3 z 4 x 4 x 3 z 4 y x y and 4 x 3 z 4 x z 34 x xyz 3 34 x3 3 x 41 3 Answer x 41 3 y 41 3 z 3 4 2 3 w 9 42 3 Reason for Lagrange using two dimensional picture Problem minimize w f x y subject to constraint g x y c Follow the level curves of f the last one to touch g c is the maximum or minimum and it is tangent gradients are parallel Reason for Lagrange using analysis Constaint g x y z c is a level surface with normal g Suppose P0 is a minimum for f on the surface Let r t be any curve on the surface with r 0 P0 h t f r t has a minimum at t 0 Taking a derivative h0 t f r t r0 t 0 h0 0 f P0 r0 0 f P0 is perpendicular to any curve on the surface through P0 f P0 is perpendicular to the surface f P0 is parallel to g P0 continued F f g z c3 z c2 z c1 18 02A topic 29 3 Example checking the boundary A rectangle in the plane is placed in the first quadrant so that one corner Q is at the origin and the two sides adjacent to Q are on the axes The corner P opposite Q is on the curve x 2y 1 Using Lagrange multipliers find for which point P the rectangle has maximum area Say how you know this point gives the maximum answer We need some names Let g x y x 2y 1 constraint and f x y xy area Gradients g bi 2bj f y bi x bj y Lagrange multipliers y x 2 OOOO OOO x 2y 1 O OO OOO The first two equations x 2y OOO x Combine this with the third equation 4y 1 y 1 4 x 1 2 P 1 2 1 4 We know this is a maximum because the maximum occurs either at a critical point or on the boundary In this case the boundary points are on the axes which gives a rectangle with area 0 Example boundary at A rectangle in the plane is placed in the first quadrant so that one corner Q is at the origin and the two sides adjacent to Q are on the axes The corner P opposite Q is on the curve xy 1 Using Lagrange multipliers find for which point P the rectangle has minimum perimeter Say how you know this point gives the minimum answer Let g x y xy 1 constraint and f x y 2x 2y perimeter Gradients g y bi x bj f 2bi 2bj y Lagrange multipliers 2 y 2 x xy 1 The first two equations x y Combine this with the third equation x2 1 x x 1 x 1 P 1 1 We know this is a minimum because the minimum occurs either at a critical point or on the boundary In this case the boundary points are infinitely far out on the axes which gives a rectangle with perimeter
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