Week 14b - QMA 2006S2 (Optimisation)

Week 14b - QMA 2006S2 (Optimisation) - Week 14 Jessica...

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Jessica Chung & Derek Hui Week 14 QMA PASS | 2006 S2 | Tues, 3-4pm | QUAD G042 Questions? Email us: [email protected] , [email protected] 1/3 U NCONSTRAINED O PTIMISATION OF F UNCTIONS OF T WO V ARIABLES Don’t be scared by the topic heading! This is very similar to previous topics of finding maxima and minima of functions except that there are now two variables to consider and so you use partial differentiation instead. Note that in this topic we are only concerned with finding local maxima extrema. For a multivariable function of two independent variables, y = f(x 1 , x 2 ) , the following conditions must be satisfied in order to determine stationary points of local maximum and minimum: First Order Conditions: The first order partial derivatives must equal to zero 1 1 0 y f x δ = = AND 2 2 0 y f x = = Second Order Conditions: Evaluate 2 11 22 12 ( ) D f f f = - to determine the nature of the stationary point i. If D > 0 but f 11 < 0 Local maximum ii. If D > 0 and f 11 > 0 Local minimum iii. If D < 0 Saddle point (neither maximum nor minimum) iv. If D = 0 Further analysis is required (test points on either side of x 1 * and x 2 *) Method for finding Local Maxima and Minima EXAMPLE: 2 3 1 2 1 1 2 2 ( , ) 2 12 3 y f x x x x x x = = - + Step 1: Find 1 y x and 2 y x . Make the equations equal to 0. 1 1 2 1 4 12 y f x x x = = - 1 2 4 12 0 x x - = (1) 2 2 1 2 2 12 9 y f x x x = = - + 2 1 2 12 9 0 x x - + = (2) Step 2: Solve for 1 1 0 y f x = = and 2 2 0 y f x = = simultaneously to obtain the stationary points (1) 1 2 4 12 0 x x - = Stationary points: (2) 2 1 2 12 9 0 x x - + = (0,0) and (12,4) Step 3: Find the second order derivatives: 2 2 1 y x , 2 2 2 y x and 2 1 2 y x x 1 1 2 1 4 12 y f x x x = = - 2 11 2 1 4 y f x = = 2 2 1 2 2 12 9 y f x x x = = - + 2 22 2 2 2 18 y f x x = = 1 1 2 1 4 12 y f x x x = = - 2 12 21 1 2 12 y f f x x = = = - 2 11 22 12 ( ) D f f f = - 2 2 2 4(18 ) ( 12) 72 144 D x x = - - = - Test stationary point at (0,0) 2 4(18 0) ( 12) (72 0) 144 144 D = × - - = × - = - As D < 0, a saddle point exists at (0,0). Step 4:
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Week 14b - QMA 2006S2 (Optimisation) - Week 14 Jessica...

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