Respuesta :
a.) P(t) = 130t - 0.4t^4 + 1200
The population is maximum when P'(t) = 0
P'(t) = 130 - 1.6t^3 = 0
1.6t^3 = 130
t^3 = 81.25
t = ∛81.25 = 4.3 months.
Maximum population P(t)max = 130(4.3) - 0.4(4.3)^4 + 1200 = 1,622
b.) The rabbit population will disappear when P(t) = 0
P(t) = 130t - 0.4t^4 + 1200 = 0
t ≈ 8.7 months
The population is maximum when P'(t) = 0
P'(t) = 130 - 1.6t^3 = 0
1.6t^3 = 130
t^3 = 81.25
t = ∛81.25 = 4.3 months.
Maximum population P(t)max = 130(4.3) - 0.4(4.3)^4 + 1200 = 1,622
b.) The rabbit population will disappear when P(t) = 0
P(t) = 130t - 0.4t^4 + 1200 = 0
t ≈ 8.7 months
Answer:
Step-by-step explanation:
P(t)=130t -0.4t^4 +1200
The population will be max when first differential of p(t) =0
So p'(t) =130-1.6t^3=0
1.6t^3=130
t^3 =130/1.6
t^3 =81.25
t = cube root of 81.25
t =4.3 months
P(max) =130(4.3) -0.4(4.3)^4 +1200
= 559-136.75+1200
=1622
Population will disappear when p(t) =0