The initial mass of a certain species of fish is 2 million tons. The mass of fish, if left alone, would increase at a rate proportional to the mass, with a proportionality constant of 4/yr. However, commercial fishing removes fish mass at a constant rate of 11 million tons per year. When will all the fish be gone? If the fishing rate is changed so that the mass of fish remains constant, what should that rate be?

Step-by-step explanation:

Let    be the amount of Radium  at any time .

Since the mass of fish would increase at a rate proportional to the mass with the proportionality constant of 4/yr, we obtain that the time rate of change is  proportional to . Therefore, in that case, the model would be an first order ODE:

However, commercial fishing removes fish mass at a constant rate of 11 million tons per year, which means that the equation is

The initial amount of fish was 2 tons, which gives the initial condition

Hence, we obtain the initial value problem

This is the mathematical model of this process.

Let's solve the ODE. Rearranging terms in the equation gives

This is an linear ODE  of the form

where   .

Hence,

So, an integrating factor is

and the general solution is given by

Now, we need to determine . To do so, we use the initial condition.

Substituting for  and for gives

Hence, the particular solution of this problem is:

Let be the moment when all fish are gone. Then, since the mass of fish is denoted by ,

We need to solve this equation for .

Let the fishing rate be such that the mass of fish remains constant.

Then, we have

Therefore,

For ,

16

step-by-step explanation:

2 x 40 = 80

80/5

= 16