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 16 16.1 The area and volume can be computed as 22Arrh 2Vrh An Excel spreadsheet can be set up to solve the problem as The formulas in the cells are: B4: =pi()*b1^2*b2 B7: =2*pi()*b1*b2 B8: =pi()*b1^2 B9: =sum(b7:b8) The Solver can be called and set up as The resulting solution is2 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. Thus, the Solver says that the optimal cylindrical container is one where the radius equals the height. For the case of the desired V = 0.5 m3, the dimensions are r = h = 0.542 m. The general result of r = h can be verified using calculus as follows. First, we can solve the volume equation for h as 2Vhr (3) This can be substituted into the area equation to give 22222VVAr r rrr We can differentiate this equation with respect to r to yield 222dA Vrdrr which can be set equal to zero and solved for 3Vr This result can then be substituted into Eq. 3 which can be solved for 3Vh Thus, we prove that the optimal container has r = h = (V/)1/3. For our desired volume of 0.5 m3, this means that r = h = (0.5/)1/3 = 0.541926 m, which confirms the result obtained numerically with the Excel Solver. 16.2 (a) The area and volume can be computed as 222Arrrh 23rhV An Excel spreadsheet can be set up to solve the problem as The formulas in the cells are: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. B4: =pi()*b1^2*b2/3 B7: =2*pi()*b1^2 B8: =pi()*b1*sqrt(b1^2+b2^2) B9: =sum(b7:b8) The Solver can be called and set up as The resulting solution is (b) For this case, the area and volume can be computed as 22Arr h 23rhV An Excel spreadsheet can be set up to solve the problem in a similar fashion to part (a) with the result: r = 0.69587 m and h = 0.98601 m. 16.3 The total area and volumes can be computed as 2222( )ArL h r 222(3)26hh rVrL (a) An Excel spreadsheet can be set up to solve the problem as4 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. The Solver can be called and set up as Note that you should also select: Options Assume non-negative. The resulting solution is Thus, the result indicates that the tank with the minimum area is spherical. (b) The constraint can be entered as5 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. The resulting solution is 16.4 This problem can be solved in a number of different ways. For example, using the golden section search, the result is i cl g(cl) c2 g(c2) c1 g(c1) cu g(cu) d copt a 1 0.0000 0.0000 3.8197 0.2330 6.1803 0.1310 10.0000 0.0641 6.1803 3.8197 100.00% 2 0.0000 0.0000 2.3607 0.3350 3.8197 0.2330 6.1803 0.1310 3.8197 2.3607 100.00% 3 0.0000 0.0000 1.4590 0.3686 2.3607 0.3350 3.8197 0.2330 2.3607 1.4590 100.00% 4 0.0000 0.0000 0.9017 0.3174 1.4590 0.3686 2.3607 0.3350 1.4590 1.4590 61.80% 5 0.9017 0.3174 1.4590 0.3686 1.8034 0.3655 2.3607 0.3350 0.9017 1.4590 38.20% 6 0.9017 0.3174 1.2461 0.3593 1.4590 0.3686 1.8034 0.3655 0.5573 1.4590 23.61% 7 1.2461 0.3593 1.4590 0.3686 1.5905 0.3696 1.8034 0.3655 0.3444 1.5905 13.38% 8 1.4590 0.3686 1.5905 0.3696 1.6718 0.3688 1.8034 0.3655 0.2129 1.5905 8.27% 9 1.4590 0.3686 1.5403 0.3696 1.5905 0.3696 1.6718 0.3688 0.1316 1.5905 5.11% 10 1.5403 0.3696 1.5905 0.3696 1.6216 0.3694 1.6718 0.3688 0.0813 1.5905 3.16% 11 1.5403 0.3696 1.5713 0.3696 1.5905 0.3696 1.6216 0.3694 0.0502 1.5713 1.98% 12 1.5403 0.3696 1.5595 0.3696 1.5713 0.3696 1.5905 0.3696 0.0311 1.5713 1.22% 13 1.5595 0.3696 1.5713 0.3696 1.5787 0.3696 1.5905 0.3696 0.0192 1.5713 0.75% Thus, after 13 iterations, the method is converging on the true value of c = 1.5679 which corresponds to a maximum specific growth rate of g = 0.36963. 16.5 (a) The LP formulation is given by Maximize 30 30 35CXYZ {Maximize profit} subject to 7 5 13 3000XY Z {Raw chemical constraint} 0.05 0.1 0.2 55XY Z {Time constraint} 450XYZ {Storage constraint}6 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
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