Zadanie 2h, Paper 3, maj 2023, TZ1, Matematyka AA IB HL
Wzorowane na: Matematyka AA IB HL — maj 2023, TZ1, Paper 3 (zadania wzorowane), pytanie 2(h). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
A design uses n positive dimensionless parameters x_1,…,x_n. Each parameter consumes one unit of setup resource plus x_j² units of variable resource. The total budget is . For S>1 and positive integer n<S, let M_n(S) be the largest achievable product x_1⋯x_n, and let P(S) be the largest M_n(S) over all admissible integers n. Initially take n=2 and S=18. You may use AM–GM: for positive z_j, . For continuous comparison define g(x)>0 by ln g(x) = (x/2)ln(S/x − 1) on 0<x<S.
Polecenie
Show that the maximum of g occurs at x=λS for a constant λ independent of S, and determine λ numerically.
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Odpowiedź z klucza
Differentiate ln g = (x/2)ln(S/x − 1) to get . Put r=S/x−1>0; the stationary equation becomes ln r=1+1/r. The function ln r−1−1/r has derivative 1/r+1/r²>0 and ranges from negative to positive values, so it has one root r≈3.5911214767. Thus x = S/(1 + r) = λS with λ≈0.2178117057, or 0.218. Since r decreases with x, g′ changes from positive to negative at this point, confirming a maximum.
Schemat punktowania
M1:prepare the logarithmic expression for differentiation or make g explicit. M1:implicit/exponential chain differentiation. M1:product and inner chain rules. A1:correct derivative relation. M1:set it to zero and reduce to an equation independent of S using x/S or r=S/x−1. A1:x=λS with λ≈0.218 and the maximum identified. Equivalent direct numerical solution for λ in 0<λ<1 is valid.
Komentarz. Paper 3: calculator and AA HL formula booklet allowed. Show proof steps. Give exact expressions where requested and otherwise three significant figures.