Mathematics, 24.03.2020 22:03, MrTeriffic

The initial mass of a certain species of fish is 33 million tons. The mass of fish, if left alone, would increase at a rate proportional to the mass, with a proportionality constant of 44/yr. However, commercial fishing removes fish mass at a constant rate of 1515 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?

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answered: fastpitchhailey1354

(a) M(t)=33e^{44\cdot 10^{-6}t}-15t

(b) t=2.203 years

Step-by-step explanation:

Modeling with Functions

We have a situation where data is managed in an effort to explain the behavior of the variables involved. It's crucial to find a mathematical expression that can model the expected behavior of the control parameters, in this case, the mass of a species of fish.

(a) Modeling

The first data provided by the problem is the initial mass, which will be called Mo=33 million tons. We also know thee mass of fish increases at a rate proportional to the existing mass. That suggests an exponential relation over time:

$M'(t)=Ce^{kt}$

Where C and k are constants, and M' is the preliminary model. Now we must consider that commercial fishing removes fish mass at a constant rate of 15 million tons per year. This means that we have to subtract this amount each year, thus the final model is:

$M(t)=Ce^{kt}-15t$

Note the mass is expressed in million tons and the time in years. The constant rate of increase is k=44/yr, this allows us to improve the model. We must be careful to express k in the proper units. The problem does not give the complete units of k, so we assume it's tons/yr, and the real value of k is

$k=44\cdot 10^{-6}$

Thus,

$M(t)=Ce^{44\cdot 10^{-6}t}-15t$

To complete the model, we use the initial condition: when t=0, M=33 million tons:

$33=Ce^{44\cdot 10^{-6}(0)}-15(0)$

This gives us

C=33

The final model is

$M(t)=33e^{44\cdot 10^{-6}t}-15t$

(b) We need to find the time t when the mass of fish is zero:

$33e^{44\cdot 10^{-6}t}-15t=0$

To solve this equation, we must use approximations since there is no way to find the value of t directly.

After some iterations, we found at t=2.203 years, the mass will go to almost zero.

(c) To save the fish from exhaustion, we must find an appropriate rate of fishing so the mass is constant. Let's call N the rate of fishing and plug it into the model:

$M(t)=33e^{44\cdot 10^{-6}t}-Nt$