ISM_T11_C04_G - Section 4.5 Applied Optimization Problems...

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Section 4.5 Applied Optimization Problems 249 (b) We require x , x , and x . Combining these requirements, the domain is the interval . ± ! # ² ") # ² #% !ß * ab (c) The maximum volume is approximately 1309.95 in. when x 3 3 in. $ ¸Þ* (d) V x x x x x . The critical point is at x w# # "% „ ±"% ± % " $' #" # "% „ &# a b œ #% ³ $$' ´ )'% œ #% ³ "% ´ $' œ œ É ab a b a b È # , that is, x or x . We discard the larger value because it is not in the domain. Since œ ( „ "$ ¸ $Þ$* ¸ "!Þ'" È V x x which is negative when x , the critical point corresponds to the maximum volume. The ww a b œ #% # ³ "% ¸ $Þ$* maximum value occurs at x , which confirms the results in (c). œ ( ³ "$ ¸ $Þ$* È (e) x x x 8 x x x x x x . Since is not in ) ³ "') ´ )'# œ ""#! Ê ³ #" ´ "!) ³ "%! œ ! Ê ) ³ # ³ & ³ "% œ ! "% $# $ # a b a b the fomain, the possible values of x are x in. or x in. œ# œ& (f) The dimensions of the resulting box are x in., x in., and x . Each of these measurements must be # #%³# ")³# positive, so that gives the domain of . !ß * 18. If the upper right vertex of the rectangle is located at x cos x for x , then the rectangle has width x and ß % !Þ& ! ² ² # 1 height cos x, so the area is A x x cos x. Solving A x graphically for x , we find that %! Þ & œ ) ! Þ & œ ! ! ² ² w 1 x . Evaluating x and cos x for x , the dimensions of the rectangle are approximately (width) by ¸ #Þ#"% # % !Þ& ¸ #Þ#"% %Þ%$ (height), and the maximum area is approximately . "Þ(* (Þ*#$ 19. Let the radius of the cylinder be r cm, r . Then the height is r and the volume is ! ² ² "! # "!! ³ È # V r r r cm . Then, V r r r r r a b a b ÈÈ Š‹ Š œ # "!! ³ œ # ³# ´ # "!! ³ # 11 1 #$ w # # # " "!! ± È r # . The critical point for r occurs at r . Since V r for œ œ !² ²"! œ œ"! ±! ±# ² % "!! ± # #!! ± $ "!! ± "!! ± #!! # $$ w 1 rr r r r # ## a b ÉÉ r and V r for r , the critical point corresponds to the maximum volume. The ²! "! ² ²"! w dimensions are r cm and h cm, and the volume is cm . œ "! ¸ )Þ"' œ ¸ ""Þ&& ¸ #%")Þ%! É # #! %!!! $ $ $ 1 20. (a) From the diagram we have 4x 108 and V x . ´jœ œ j # The volume of the box is V(x) x (108 4x), where œ³ # 0 x 27. Then Ÿ² V (x) 216x 12x 12x(18 x) 0 œ³œ ³ œ the critical points are 0 and 18, but x 0 results in Êœ no box. Since V (x) 216 24x 0 at x 18 we ww œ³² œ have a maximum. The dimensions of the box are 18 18 36 in. ‚‚
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250 Chapter 4 Applications of Derivatives (b) In terms of length, V( ) x . The graph jœ jœ j # ±j # ˆ‰ 108 4 indicates that the maximum volume occurs near 36, which is consistent with the result of part (a). 21. (a) From the diagram we have 3h 2w 108 and ±œ V h w V(h) h 54 h 54h h . œÊ œ ² œ² ## # $ 33 Then V (h) 108h h h(24 h) 0 w# œ ² œ 99 h 0 or h 24, but h 0 results in no box. Since Êœ œ œ V (h) 108 9h 0 at h 24, we have a maximum ww ³ œ volume at h 24 and w 54 h 18. œœ ² œ 3 # (b) 22. From the diagram the perimeter is P 2r 2h r, œ±± 1 where r is the radius of the semicircle and h is the height of the rectangle. The amount of light transmitted proportional to A 2rh r r(P 2r r) r œ± ²± "" 44 11 1 rP 2r r . Then P 4r r 0 œ² ² œ²² œ # 3d A 3 4d r r 2h P .
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This note was uploaded on 09/20/2010 for the course MATHEMATIC 09991051 taught by Professor Dr.maenshadeed during the Fall '10 term at Norwegian Univ. of Science & Technology.

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ISM_T11_C04_G - Section 4.5 Applied Optimization Problems...

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