1 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. CHAPTER 14 14.1 The partial derivatives can be evaluated, 22(2)4fxx 44(2)8fyy The angle in the direction of h is 13tan 0.982794 radians ( 56.30993 )2 The directional derivative can be computed as '(0) 4cos(0.982794) 8sin(0.982794) 8.875203g 14.2 The elevation can be determined as 22(0.8, 1.2) 2(0.8)1.2 1.5(1.2) 1.25(0.8) 2(1.2) 5 5.04f The partial derivatives can be evaluated, 2 2.5 2(1.2) 2.5(0.8) 0.4fyxx 2 1.5 4 2(0.8) 1.5 4(1.2) 1.7fxyy which can be used to determine the gradient as f = 0.4i – 1.7j. This corresponds to the direction = tan–1(–1.7/0.4) = –1.3397 radians (= –76.76o). This vector can be sketched on a topographical map of the function as shown below: 0 0.4 0.8 1.2 1.6 200.40.81.21.62 The slope in this direction can be computed as 220.4 ( 1.7) 1.746 2 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 14.3 The partial derivatives can be evaluated, 32.25fxyx 2.25 4 1.75fxyy These can be set to zero to generate the following simultaneous equations 32.25 0xy 2.25 4 1.75xy which can be solved for x = 0.567568 and y = 0.756757, which is the optimal solution. 14.4 The partial derivatives can be evaluated at the initial guesses, x = 1 and y = 1, 3 2.25 3(1) 2.25(1) 0.75fxyx 2.25 4 1.75 2.25(1) 4(1) 1.75 0fxyy Therefore, the search direction is –0.75i. 2(1 0.75 , 1) 0.5 0.5625 0.84375fhhh This can be differentiated and set equal to zero and solved for h* = 0.33333. Therefore, the result for the first iteration is x = 1 – 0.75(0.3333) = 0.75 and y = 1 + 0(0.3333) = 1. For the second iteration, the partial derivatives can be evaluated as, 3(0.75) 2.25(1) 0fx 2.25(0.75) 4(1) 1.75 0.5625fy Therefore, the search direction is –0.5625j. 2(0.75, 1 0.5625 ) 0.59375 0.316406 0.63281fhhh This can be differentiated and set equal to zero and solved for h* = 0.25. Therefore, the result for the second iteration is x = 0.75 + 0(0.25) = 0.75 and y = 1 + (–0.5625)0.25 = 0.859375.3 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 0 0.2 0.4 0.6 0.8 1 1.200.20.40.60.811.2021max 14.5 (a) 22343xyxyyyefxy xe 22343343 3 43xyxyxyxy xy xyy e y xye eHy xye e x x e (b) 224xfyz 200020004H (c) 222222232623xyxxy yfxyxxy y 222 2222 2222242 21262126 2121823xxy y x xy yxxy y x xy yHxxyy 14.6 The partial derivatives can be evaluated at the initial guesses, x = 1 and y = 1, 2( 3) 2(1 3) 4fxx 2( 2) 2(1 2) 2fyy 22(1 4 , 1 2 ) (1 4 3) (1 2 2)fhh h h 22() (4 2) (2 1)gh h h Setting g(h) = 0 gives h* = –0.5. Therefore, x = 1 –4(–0.5) = 3 y = 1 –2(–0.5) = 2 Thus, for this special case, the approach converges on the correct answer after a single iteration. This occurs because the function is spherical as shown below. Thus, the gradient for any guess points directly at the solution.4 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 00.81.62.43.2411.82.63.44.25 14.7 The partial derivatives can be evaluated at the initial guesses, x = 0 and y = 0, 3342 8 2 42(0)8(0) 2(0)4fxx yx 2 2 6 2 2(0) 6(0) 2fxyy f(0 + 4h, 0 + 2h) = 20h + 20h2 – 512h4 g(h) = 20 + 40h – 2048h3 The root of this equation can be determined by bisection. Using initial guesses of h = 0 and 1 yields a root of h* = 0.24390 after 13 iterations with a = 0.05%. Therefore, x = 0 + 4(0.24390) = 0.976074 y = 0 + 2(0.24390) = 0.488037 14.8 82 2fxyx 12 8 2fyxy At x = y = 0, 8fx 12fy (0 8 ,0 12 ) ( )fhhgh 2() 832 208ghhh At g(h) = 0, h* = 0.125. Therefore, x = 0 8(0.125) = 1 y = 0 + 12(0.125) = 1.5 14.9 The following code implements the random search algorithm in VBA. It is set up to solve Prob. 14.7.5 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are
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