Answer
(a) $3.23\times 10^7~kg$
(b) It takes 1.55 years to reach a biomass of $4\times 10^7~kg$
Work Step by Step
(a) A logistic equation has this form:
$\frac{dy}{dt} = ky(1-\frac{y}{M})$
The solution has this form:
$y(t) = \frac{M}{1+Ae^{-kt}}~~$ where $~~A = \frac{M-y_0}{y_0}$
We can find $A$:
$A = \frac{M-y_0}{y_0} = \frac{(8\times 10^7)-(2\times 10^7)}{2\times 10^7} = 3$
We can write the solution:
$y(t) = \frac{8\times 10^7}{1+3e^{-0.71t}}$
We can find the biomass after 1 year:
$y(t) = \frac{8\times 10^7}{1+3e^{(-0.71)(1)}} = 3.23\times 10^7~kg$
(b) We can find the time $t$ it takes to reach a biomass of $4 \times 10^7~kg$:
$y(t) = \frac{8\times 10^7}{1+3e^{-0.71t}} = 4\times 10^7$
$1+3e^{-0.71t} = \frac{8\times 10^7}{4\times 10^7}$
$1+3e^{-0.71t} = 2$
$3e^{-0.71t} = 1$
$e^{-0.71t} = \frac{1}{3}$
$-0.71t = ln(\frac{1}{3})$
$t = (\frac{1}{-0.71})~ln(\frac{1}{3})$
$t = 1.55$
It takes 1.55 years to reach a biomass of $4\times 10^7~kg$
