Answer:
a) The rate of production of fishers per month is changing with time at a rate of 40.5 in 4 months from the start of operations.
b) In 4 months from now, the fisher's production will be 1081.53 pounds of fishes per month.
Step-by-step explanation:
The calculation is presented in the attached images to this solution.
Therefore, the fisher's production will be an increase [tex](\frac{dQ}{dt} > 0)[/tex] at the rate of [tex]42.09[/tex] pounds per month.
The rate of change of one quantity with respect to another is one of the major applications of derivatives.
Given function is,
[tex]Q(k)=111 k^{\frac{1}{3} }[/tex]
And [tex]k(t)=0.5t^2+100t+517[/tex]
The rate of production of fish changing with respect to time [tex]t[/tex] then,
[tex]\frac{dQ}{dt}= \frac{dQ}{dk} \times \frac{dk}{dt}\\=\frac{111}{3} k^{\frac{-2}{3} }(t+100)[/tex]
When [tex]t=4[/tex]
[tex]\frac{dQ}{dt} =3996k^{-\frac{2}{3} }...(1)[/tex]
Here [tex]k[/tex] is the function of [tex]t[/tex]
Calculating [tex]k[/tex]
[tex]k(4)=0.5(4)^2+100(4)+517\\=925[/tex]
Now, substituting in the equation (1)
[tex]\frac{dQ}{dt} =3996(925)^{-\frac{2}{3}}\\ =42.09[/tex]
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