Pre-Calc Homework Solutions 164

# Pre-Calc Homework Solutions 164 - 164 Section 4.4 46. 4.25,...

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43. continued (b) Note that x . 4.25, and let f ( x ) 5 L 2 5 } 2 x 2 2 x 3 8.5 } . Since y # 11, the approximate domain of f is 5.20 # x # 8.5. Then f 9 ( x ) 55 For x . 5.20, the critical point occurs at x 5 } 5 8 1 } 5 6.375 in., and this corresponds to a minimum value of f ( x ) because f 9 ( x ) , 0 for 5.20 , x , 6.375 and f 9 ( x ) . 0 for x . 6.375. Therefore, the value of x that minimizes L 2 is x 5 6.375 in. (c) The minimum value of L is ! } 2 § (6 § 2 .3 § (6 7 § . 5 3 § ) 7 § 2 5 § ) 3 § 8 § .5 } § < 11.04 in. 44. Since R 5 M 2 1 } C 2 } 2 } M 3 } 2 5 } C 2 } M 2 2 } 1 3 } M 3 , we have } d d M R } 5 CM 2 M 2 . Let f ( M ) 5 CM 2 M 2 . Then f 9 ( M ) 5 C 2 2 M , and the critical point for f occurs at M 5 } C 2 } .This value corresponds to a maximum because f 9 ( M ) . 0 for M , } C 2 } and f 9 ( M ) , 0 for M . } C 2 } . The value of M that maximizes } d d M R } is M 5 } C 2 } . 45. The profit is given by P ( x ) 5 ( n )( x 2 c ) 5 a 1 b (100 2 x )( x 2 c ) 52 bx 2 1 (100 1 c ) bx 1 ( a 2 100 bc ). Then P 9 ( x ) 2 bx 1 (100 1 c ) b 5 b (100 1 c 2 2 x ). The critical point occurs at x 5 } 100 2 1 c } 5 50 1 } 2 c } , and this value corresponds to the maximum profit because P 9 ( x ) . 0 for x , 50 1 } 2 c } and P 9 ( x ) , 0 for x . 50 1 } 2 c } .
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## This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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