幾題商管類微積分考題

Q1:
f(x)=x^a (1-x)^b, 0<= x <= 1
f(0)=0=f(1) , 而f(x)>=0, so, 求max. 可設 0<x<1
令 g(x)=ln[f(x)]= a ln(x)+ b ln(1-x) (求 g(x)之 max.即可得 f(x)之max.)
g'(x)= a/x - b/(1-x) = [a- (a+b)x]/[x(1-x)]
so, 0<x<1時, 當 x= a/(a+b)時, g'(x)=0,
x<a/(a+b)時 g'(x)>0, x>a/(a+b)時g'(x)<0
thus, x=a/(a+b)時 g(x)最大, f(x)亦最大,
hence, max. f(x) = f[a/(a+b)]= (a^a)*(b^b)/[(a+b)^(a+b)]

Q2:
設起點為(0,0),A往東,B往北,第t小時的位置A(3t,0), B(0, 2t)
15分鐘時, t=1/4 小時,
AB距離f(t)= √[(3t)^2+(2t)^2]=√13 t
so,AB距離的變化率=df/dt = √13 (km/hr) (任何時間均相同)

Q3:
(b)可解過, T=20
(c) C(t)=(1/t)∫[0~t] [f(s)+g(s)] ds [求t=0~20時, C(t)最小值]
=(V/t) ∫[0~ t] [(1/10)-(s/200)+(s^2 / 12000)] ds
=(V/t) [ (t/10)-(t^2)/ 400 + t^3/36000 ]=V[ (1/10)- t/400+ t^2/36000]
C'(t)= V[-1/400+ t/18000]
so, C'(t)<0 for 0<t<=20, 故 C(t)遞減
C(20)=V[(1/10)- 20/400+ 400/36000]= 11V/180為絕對最小值

Q4:
dP/dt = kP(N-P), P(0)=2
∫dP/[P(N-P)] = ∫k dt
∫[1/P + 1/(N-P) ]dP = ∫Nk dt
ln| P/(N-P) | = Nk t + c
P/(N-P)= (1/C) exp(Nk t),
so, P= N exp(Nk t)/[ C+ exp(Nk t) ]= N/[C exp(-Nk t) +1]
P(0)=2, then 2= N/(C+1), C=N/2 - 1
thus, P= N/[(N/2 -1) exp(-Nk t) +1] ----(*)
顯然P遞增,反曲點時, dP/dt=kP(N-P)最大
kP(N-P)= -k[(P-N/2)^2 - N^2/4]= -k(P-N/2)^2 + kN2/4
故P=N/2時為反曲點,代入(*)得
N/2 = N/[ (N/2 -1) exp(-Nk t)+1]
(N/2 -1) exp(-Nk t) +1 =2,
exp(Nk t) = (N/2 -1)
so, t= ln(N/2 -1)/(Nk)
Ans: 反曲點 t=ln(N/2 -1)/(Nk), P=N/2

maximum value of f(x) = x^a(1-x)^b , and 0 <= x <= 1:

a^a * b^b / (a + b)^(a + b)