Dom, the manager of a fishery, determines that t weeks after 300 fish of a parti

Question

Dom, the manager of a fishery, determines that t weeks after 300 fish of a particular species are released into a pond the average weight of an individual fish (in pounds) for the first 10 weeks will be w (t) = 3 + t - 0.05t^2 Dom further determines that the proportion of the fish that are still alive after t weeks is given by P (t) = 31/31 + t The expected yield (Y (t) of the fish after t weeks is the total weight of the fish that are still alive. Express Y (t) in terms of w (t) and p (t) for 0 lessthanorequalto t lessthanorequalto 10/Y (t) = 300 * w (t) p (t) = When should Dom expect the yield Y (*t) to be maximized? What is the maximum yield? Y'(t) = The denominator of Y'(t) is always > 0. Disregard the 9300 and solve quadratic in the numerator for t = 0. Disregard the negative t. t = Y =

Explanation / Answer

given w(t)=3+t-0.05t2, p(t)=31/(31+t)

a)Y(t)=300*w(t)p(t)

Y(t)=300(3+t-0.05t2)(31/(31+t))

Y(t)=9300((3+t-0.05t2)/(31+t))

b)Y'(t)=9300([(0+1-0.1t)(31+t) -(3+t-0.05t2)(0+1)]/(31+t)2)

Y'(t)=9300([(1-0.1t)(31+t) -(3+t-0.05t2)]/(31+t)2)

Y'(t)=9300([31-3.1t+t-0.1t2 -3-t+0.05t2]/(31+t)2)

Y'(t)=9300([28-3.1t -0.05t2]/(31+t)2)

Y'(t)=0

9300([28-3.1t -0.05t2]/(31+t)2)=0

[28-3.1t -0.05t2]=0

t=[3.1+((-3.1)2-4*(-0.05)*28))]/(2*-0.05),t=[3.1-((-3.1)2-4*(-0.05)*28))]/(2*-0.05)

t=-70, t=8

t=8

Y(8)=9300((3+8-0.0582)/(31+8))

Y(8)=1860

dom expect yield to be maximised at t=8

maximum yield =1860

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