how to solve this question ( calculus ,integration )?!

Question

anonymous · Answer

attachment

anonymous · Answer

$$f'(x)=60{x}^{1/2}$$ after integration $$f(x)=40{x}^{2/3}+c$$

anonymous · Answer

@satellite73  @ganeshie8  @RadEn  @Callisto  please help solve this question .

raden · Answer

that should 40x^(3/2) , not 40x^2/3

anonymous · Answer

oky my bad fixd ,what is the next step? @RadEn  $$f(x)=40{x}^{3/2} $$

ranga · Answer

f'(x) = 60x^1/2
f(x) = 60 * x^3/2 * 2/3 + C
f(x) = 40 * x^3/2 + C
At x = 0, f(x) = 27,000
27,000 = C
f(x) = 40 * x^3/2 + 27000
Find x when f(x) = 41,000
41,000 = 40 * x^3/2 + 27000
solve for x

anonymous · Answer

wow is it that simple $$\frac{ (41,000-27000)}{ 40 }  =  x ^{3/2} $$ == $$\frac{ (14000)}{ 40 }  =  x ^{3/2} $$ $$350  =  x ^{3/2} $$  ==$$\ln 350=\ln  x ^{3/2} $$ $$\ln 350=\frac{ 3 }{ 2 } lnx  $$ =$$\frac{ 5.857933154}{ 3/2 }=lnx  $$  ==$$lnx  =3.90528877$$ ==$$x= e^{3.90528877}$$  x= 49.6644 is it true @ranga

ranga · Answer

Yes.  I would round it up to 50 days as the answer.

anonymous · Answer

Thank you so much for the help @ranga

ranga · Answer

You are welcome.