43. continued(b)Note that x .4.25, and let f(x)5L25}2x22x38.5}. Sincey#11, the approximate domain of fis 5.20 #x #8.5.Thenf9(x)55For x .5.20, the critical point occurs at x5 }581} 56.375 in., and this corresponds to a minimumvalue of f(x) because f9(x) ,0 for 5.20 ,x ,6.375and f9(x) .0 for x .6.375. Therefore, the value of xthat minimizes L2is x56.375 in.(c)The minimum value of Lis !}2§(6§2.3§(67§.53§)7§25§)3§8§.5}§<11.04 in.44.Since R5M21}C2}2 }M3}25 }C2}M22 }13}M3, we have }ddMR} 5CM2M2. Let f(M)5CM2M2. Then f9(M)5C22M, and the critical point for foccurs atM5 }C2}.This value corresponds to a maximum becausef9(M) .0 for M , }C2}and f9(M) ,0 for M . }C2}. The valueof Mthat maximizes }ddMR}is M5 }C2}.45.The profit is given byP(x)5(n)(x2c)5a1b(1002x)(x2c)52bx21(1001c)bx1(a2100bc).Then P9(x)2bx1(1001c)b5b(1001c22x).The critical point occurs at x5}10021c} 5501 }2c}, and thisvalue corresponds to the maximum profit because P9(x) .0 for x ,501 }2c}and P9(x) ,0 for x .501 }2c}.
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