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 13 13.1 (a) The function can be differentiated to give '( ) 2 8fxx This function can be set equal to zero and solved for x = 8/2 = 4. The derivative can be differentiated to give the second derivative "( ) 2fx Because this is negative, it indicates that the function has a maximum at x = 4. (b) Using Eq. 13.7 x0 = 0 f(x0) = –12 x1 = 2 f(x1) = 0 x2 = 6 f(x2) = 0 312(4 36) 0(36 0) 0(0 4)42( 12)(2 6) 2(0)(6 0) 2(0)(0 2)x 13.2 (a) The function can be plotted -120-80-40040-2 -1 0 1 2 (b) The function can be differentiated twice to give 42"( ) 45 24fxxx Thus, the second derivative will always be negative and hence the function is concave for all values of x. (c) Differentiating the function and setting the result equal to zero results in the following roots problem to locate the maximum 53'( ) 0 9 8 12fx x x A plot of this function can be developed2 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. -400-2000200400-2 -1 0 1 2 A technique such as bisection can be employed to determine the root. Here are the first few iterations: iteration xl xu xr f(xl) f(xr) f(xl)f(xr) a 1 0.00000 2.00000 1.00000 12 -5 -60.0000 2 0.00000 1.00000 0.50000 12 10.71875 128.6250 100.00% 3 0.50000 1.00000 0.75000 10.71875 6.489258 69.5567 33.33% 4 0.75000 1.00000 0.87500 6.489258 2.024445 13.1371 14.29% 5 0.87500 1.00000 0.93750 2.024445 -1.10956 -2.2463 6.67% The approach can be continued to yield a result of x = 0.91692. 13.3 First, the golden ratio can be used to create the interior points, 51(2 0) 1.23612d 10 1.2361 1.2361x 22 1.2361 0.7639x The function can be evaluated at the interior points 2( ) (0.7639) 8.1879fx f 1( ) (1.2361) 4.8142fx f Because f(x2) > f(x1), the maximum is in the interval defined by xl, x2, and x1.where x2 is the optimum. The error at this point can be computed as 20(1 0.61803) 100% 100%0.7639a For the second iteration, xl = 0 and xu = 1.2361. The former x2 value becomes the new x1, that is, x1 = 0.7639 and f(x1) = 8.1879. The new values of d and x2 can be computed as 51(1.2361 0) 0.76392d 21.2361 0.7639 0.4721x The function evaluation at f(x2) = 5.5496. Since this value is less than the function value at x1, the maximum is in the interval prescribed by x2, x1 and xu. The process can be repeated and all three iterations summarized as3 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. i xl f(xl) x2 f(x2) x1 f(x1) xu f(xu) d xopt a 1 0.0000 0.0000 0.7639 8.1879 1.2361 4.8142 2.0000 -104.0000 1.2361 0.7639 100.00% 2 0.0000 0.0000 0.4721 5.5496 0.7639 8.1879 1.2361 4.8142 0.7639 0.7639 61.80% 3 0.4721 5.5496 0.7639 8.1879 0.9443 8.6778 1.2361 4.8142 0.4721 0.9443 30.90% 13.4 First, the function values at the initial values can be evaluated 0() (0)0fx f 1() (1)8.5fx f 2() (2) 104fx f and substituted into Eq. (13.7) to give, 22 22 2230(1 2 ) 8.5(2 0 ) ( 104)(0 1 )0.5702482(0)(1 2) 2(8.5)(2 0) 2( 104)(0 1)x which has a function value of f(0.570248) = 6.5799. Because the function value for the new point is lower than for the intermediate point (x1) and the new x value is to the left of the intermediate point, the lower guess (x0) is discarded. Therefore, for the next iteration, 0( ) (0.570248) 6.6799fx f 1() (1)8.5fx f 2() (2) 104fx f which can be substituted into Eq. (13.7) to give x3 = 0.812431, which has a function value of f(0.812431) = 8.446523. At this point, an approximate error can be computed as 0.81243 0.570248100% 29.81%0.81243a The process can be repeated, with the results tabulated below: i x0 f(x0) x1 f(x1) x2 f(x2) x3 f(x3) a 1 0.00000 0.00000 1.00000 8.50000 2.0000 -104 0.57025 6.57991 2 0.57025 6.57991 1.00000 8.50000 2.0000 -104 0.81243 8.44652 29.81% 3 0.81243 8.44652 1.00000 8.50000 2.0000 -104 0.90772 8.69575 10.50% Thus, after 3 iterations, the result is converging on the true value of f(x) = 8.69793 at x = 0.91692. 13.5 The first and second derivatives of the function can be evaluated as 53'( ) 9 8 12fx x x 42"( ) 45 24fxxx which can be substituted into Eq. (13.8) to give 53142981245 24iiiiiixxxxxx4 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. Substituting the initial guess yields 531429(2 ) 8(2 ) 12 3402 2 1.58333381645(2 ) 24(2 )ix which has a function value of –17.2029. The second iteration gives 531429(1.583333 ) 8(1.583333 ) 12 109.3131.583333 1.583333 1.26462342.98145(1.583333 ) 24(1.583333 )ix which has a function value of 3.924617. At
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