Question #320390

It is estimated that x months from now the population of a certain town will be increasing at the rate of 2 + 6√𝑥 people per month. The current population is 5,000. What will be the population 9 months from now?


1
Expert's answer
2022-04-06T07:37:50-0400

Let


dpdx=2+6x\frac{dp}{dx}=2+6\sqrt{x}


dp=(2+6x)dxdp=(2+6\sqrt{x})dx


Integrating both sides, we have that


P(x)=2x+4x32+kP(x)=2x+4x^{\frac{3}{2}}+k


at x=0,P=5000x=0, P=5000


5000=2(0)+4(0)32+k5000=2(0)+4(0)^{\frac{3}{2}}+k


=>k=5000=>k=5000



P(x)=2x+4x32+5000P(x)=2x+4x^{\frac{3}{2}}+5000


When x=9x=9


P=2(9)+4(9)32+5000P=2(9)+4(9)^{\frac{3}{2}}+5000


P=18+108+5000=5126P=18+108+5000=5126


Hence, Population 9months from now is 5126



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