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The biomass B(t) of a fishery is the total mass of the members of the fish population at time t. It is the product of the number of individuals N(t) in the population and the average mass M(t) of a fish at time t. In the case of guppies, breeding occurs continually. Suppose that at time t = 4 weeks the population is 829 guppies and is growing at a rate of 50 guppies per week, while the average mass is 1.2 g and is increasing at a rate of 0.14 g/week.
At what rate is the biomass increasing when t = 4? (Round your answer to one decimal place.)

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gkosworldreign

Answer:

174.8 guppies.g/week

Step-by-step explanation:

B(t) = N(t)M(t)

Utilizing the product rule of derivative

[tex]\frac{d(AH)}{dt} = A\frac{dH}{dt} + H\frac{dA}{dt}[/tex]

Applying the rule to B(t) = N(t)M(t)

[tex]\frac{d(B(t))}{dt} = N(t)\frac{d(M(t))}{dt} + M(t)\frac{d(N(t))}{dt}[/tex]

from the question

t=4 weeks, N(4) = 820 guppies

d(N(4))/dt = 50 guppies/week

m(4) = 1.2g

d(M(4))/dt = 0.14g/week

[tex]\frac{d(B(t))}{dt} = N(t)\frac{d(M(t))}{dt} + M(t)\frac{d(N(t))}{dt}\\\\= 820(0.14) + 50 (1.2) = 174.8 guppies.g/week[/tex]

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