Answer:
The value of dx/dt is [tex]\frac{100p}{9\sqrt{(810000-p^2)}}.\frac{1600}{\sqrt{t}\left(\sqrt{t}+8\right)^2}\\[/tex] while the rate of change of quantity demanded per month after 16 months is 18.7.
Step-by-step explanation:
From the given data the equation of quantity demanded for average price is given as is given as
[tex]x=f(p)=\frac{100}{9}\sqrt{810000-p^2}[/tex]
The equation of average price p for a given value of t is given as
[tex]p(t)=\frac{400}{1+\frac{1}{8}\sqrt{t}}+200[/tex]
Now in order to determine the rate at which the quantity demanded will be changing is given as [tex]\frac{dx}{dt}[/tex]
This is found by using the chain rule as
[tex]\frac{dx}{dt}=\frac{dx}{dp}.\frac{dp}{dt}[/tex]
Now
[tex]\frac{dx}{dp}=\frac{d(\frac{100}{9}\sqrt{810000-p^2})}{dp}\\\frac{dx}{dp}=\frac{100}{9}\frac{d}{dp}[(810000-p^2)^{1/2}]\\\frac{dx}{dp}=\frac{100}{9}\frac{1}{2}[(810000-p^2)^{-1/2}]\frac{d}{dp}[(810000-p^2)]\\\frac{dx}{dp}=\frac{50}{9}[(810000-p^2)^{-1/2}](-2p)\\\frac{dx}{dp}=\frac{100p}{9\sqrt{(810000-p^2)}}[/tex]
Now
[tex]\\\frac{dp}{dt}=\frac{d(\frac{400}{1+\frac{1}{8}\sqrt{t}}+200)}{dt}\\\frac{dp}{dt}=\frac{d}{dt}\left(\frac{400}{1+\frac{1}{8}\sqrt{t}}+200\right)\\\frac{dp}{dt}=\frac{d}{dt}\left(\frac{400}{1+\frac{1}{8}\sqrt{t}}\right)+0\\\frac{dp}{dt}=400\frac{d}{dt}\left(\frac{1}{1+\frac{1}{8}\sqrt{t}}\right)\\\frac{dp}{dt}=400\frac{d}{du}\left(\left(1+\frac{1}{8}\sqrt{t}\right)^{-1}\right)\frac{d}{dt}\left(1+\frac{1}{8}\sqrt{t}\right)\\\frac{dp}{dt}=-\frac{1600}{\sqrt{t}\left(\sqrt{t}+8\right)^2}[/tex]
So now the value of dx/dt is given as
[tex]\frac{dx}{dt}=\frac{dx}{dp}.\frac{dp}{dt}\\\frac{dx}{dt}=\frac{100p}{9\sqrt{(810000-p^2)}}.\frac{1600}{\sqrt{t}\left(\sqrt{t}+8\right)^2}\\[/tex]
So the value of dx/dt is [tex]\frac{100p}{9\sqrt{(810000-p^2)}}.\frac{1600}{\sqrt{t}\left(\sqrt{t}+8\right)^2}\\[/tex]
Now for the time =16 months price is given as
[tex]p(t)=\frac{400}{1+\frac{1}{8}\sqrt{t}}+200\\p(16)=\frac{400}{1+\frac{1}{8}\sqrt{16}}+200\\p(16)=466.67[/tex]
Now the value of x is given as
[tex]\frac{dx}{dt}=\frac{100p}{9\sqrt{(810000-p^2)}}.\frac{1600}{\sqrt{t}\left(\sqrt{t}+8\right)^2}\\\\\frac{dx}{dt}=\frac{100*466.67}{9\sqrt{(810000-466.67^2)}}.\frac{1600}{\sqrt{16}\left(\sqrt{16}+8\right)^2}\\\frac{dx}{dt}=18.72[/tex]
So the rate of change of quantity demanded per month after 16 months is 18.72
