Science:Math Exam Resources/Courses/MATH101/April 2007/Question 06 (b)/Solution 1

We want to find the steady states of this differential equations. To do so, set ${\frac {dm}{dt}}=0$, i.e.

$16m\left(1-{\frac {m}{4}}\right)-12=0$

Simplifying yields

$16m-4m^{2}-12=0$

and once more

$m^{2}-4m+3=0$

Factoring yields

$(m-3)(m-1)=0$

and so the steady states are given by $m=1$ and $m=3$. It is quickly checked, e.g. by plugging in $m=0$, $m=2$, and $m=4$, that

${\frac {dm}{dt}}=\left\{{\begin{array}{rl}{\text{negative }}&{\text{if }}m<1{\text{ or }}m>3,\\0&{\text{if }}m=1{\text{ or }}m=3,\\{\text{positive }}&{\text{if }}1<m<3.\end{array}}\right.$

At time $t=0$ we have $m(0)=2$. Hence, despite the fishing, the population of fish will grow initially. But the growth will slow down as the population size approaches the steady state value $m=3$. A steady state can never be reached in finite time, hence the answer is No, the fish population will never equal 3 million.

Caption: We plot the rate of change $dm/dt$ ($y$-axis) as function of population size $m$ ($x$-axis).