{VERSION 4 0 "IBM INTEL NT" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Out put" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 256 99 "Here's the equation f; let's find the de rivative, set it to zero, and see the best k in terms of p." } {MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "f := ((1-p)^k + (1 - (1-p)^k)*(k+1))*(n/k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG *&*&,&),&\"\"\"F*%\"pG!\"\"%\"kGF**&,&F*F*F(F,F*,&F-F*F*F*F*F*F*%\"nGF *F*F-F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "df := diff(f,k); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#dfG,&*&*&,**&),&\"\"\"F,%\"pG! \"\"%\"kGF,-%#lnG6#F+F,F,*(F*F,F0F,,&F/F,F,F,F,F.F,F,F*F.F,%\"nGF,F,F/ F.F,*&*&,&F*F,*&,&F,F,F*F.F,F4F,F,F,F5F,F,*$)F/\"\"#F,F.F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "k = solve(df=0,k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%\"kG6$,$*&-%)LambertWG6#,$*$-%%sqrtG6#,$-%#lnG6# ,&\"\"\"F5%\"pG!\"\"F7F5#F5\"\"#F5F1F7F9,$*&-F)6#,$F,#F7F9F5F1F7F9" }} }{EXCHG {PARA 0 "" 0 "" {TEXT 259 12 "This is the " }{TEXT 261 5 "exac t" }{TEXT 262 317 " answer to what k should be. If anyone in the clas s knows how to use the Lambert W equation, he/she shouldn't be in this class. Instead, we need to find a simplification to this problem. T hrough the binomial theorem we know that (1-n)^r for small enough n is approximately (1-nr). We can use this in our equation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "g := ((1 - p*k) + (1 - (1 - p*k))*( k+1))*n/k;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gG*&*&,(\"\"\"F(*&% \"pGF(%\"kGF(!\"\"*(F*F(F+F(,&F+F(F(F(F(F(F(%\"nGF(F(F+F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "g := simplify(g);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"gG*&*&,&\"\"\"F(*&%\"pGF()%\"kG\"\"#F(F(F(%\"nGF (F(F,!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "dg := diff(g, k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#dgG,&*&%\"pG\"\"\"%\"nGF(\" \"#*&*&,&F(F(*&F'F()%\"kGF*F(F(F(F)F(F(*$F/F(!\"\"F2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "This looks like a more reasonable answer. We' ll take the positive root in this case, which is 1 / sqrt(p)." } {MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "k = solve(dg=0, k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%\"kG6$*&\"\"\"F'*$-%%sqrtG6#% \"pGF'!\"\",$F&F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 248 "Let's check our answers by plugging in values for p, n, and k. Notice how functi ons f and g give us almost exactly the same answer (which is the expec ted number of tests performed at this drug percentage, total number of people, and size of group." }{MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "p:=0.01; n:=10000; k:=1/sqrt(p);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"pG$\"\"\"!\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% \"nG\"&++\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG$\"+++++5!\")" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(f);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#$\"+]#zh&>!\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(g);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+++++?!\"'" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "4 0 2" 146 } {VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }