#### To determine

**To find:** The rate at which the biomass is increasing when t=4.

#### Answer

The biomass is increasing at a rate of B′(4)=174.8 g/week when t=4.

#### Explanation

**Given:**

The biomass B(t) of fish population is the product of the number of individuals N(t) in the population at time *t* and the average mass M(t) of a fish at a time *t*.

That is, B(t)=N(t)M(t).

The population at t=4 weeks is 820 guppies. That is, N(4)=820.

The population grows at a rate of 50 guppies per week. That is, N′(t)=50.

The average mass is 1.2 gram. That is, M(4)=1.2.

The mass is increasing at rate of 0.14 gram per week. That is, M′(t)=0.14.

**Derivative rule:**

Product Rule: ddx[f1(x)f2(x)]=f1(x)ddx[f2(x)]+f2(x)ddx[f1(x)]

**Calculation:**

Obtain the rate at which the biomass is increasing when t=4 as follows.

The derivative of B(t)=N(t)M(t) that is B′(t) is obtained as follows,

B′(t)=ddt(N(t)M(t))

Apply the Product rule (1) and simplify the terms,

B′(t)=N(t)ddt(M(t))+M(t)ddt(N(t))=N(t)M′(t)+M(t)N′(t)

Thus, the derivative of B(t)=N(t)M(t) is B′(t)=N(t)M′(t)+M(t)N′(t).

Substitute t=4 in the equation for B′(t) as,

B′(4)=N(4)M′(4)+M(4)N′(4)

Substitute the values of N(4)=820, N′(4)=50, M(4)=1.2, and M′(4)=0.14 in B′(4) as,

B′(4)=(820×0.14)+(1.2×50)=114.8+60=174.8

Therefore, the rate at which the biomass is increasing when t=4 is B′(4)=174.8 g/week.