Respuesta :
Answer:
(a)[tex]D(t)=\dfrac{50000}{1.9t+9}[/tex]
(b)[tex]D'(115)=-1.8355[/tex]
Step-by-step explanation:
The demand function for a product is given by :
[tex]D(p)=\dfrac{50000}{p}[/tex]
Price, p is a function of time given by [tex]p=1.9t+9[/tex], where t is in days.
(a)We want to find the demand as a function of time t.
[tex]\text{If } D(p)=\dfrac{50000}{p},$ and p=1.9t+9\\Then:\\D(t)=\dfrac{50000}{1.9t+9}[/tex]
(b)Rate of change of the quantity demanded when t=115 days.
[tex]\text{If } D(t)=\dfrac{50000}{1.9t+9}[/tex]
[tex]\dfrac{\mathrm{d}}{\mathrm{d}t}\left[\dfrac{50000}{\frac{19t}{10}+9}\right]}}=50000\cdot \dfrac{\mathrm{d}}{\mathrm{d}t}\left[\dfrac{1}{\frac{19t}{10}+9}\right]}[/tex]
[tex]=-50000\cdot\dfrac{d}{dt} \dfrac{\left[\frac{19t}{10}+9\right]}{\left(\frac{19t}{10}+9\right)^2}}}[/tex]
[tex]=\dfrac{-50000(1.9\frac{d}{dt}t+\frac{d}{dt}9)}{\left(\frac{19t}{10}+9\right)^2}}}[/tex]
[tex]=-\dfrac{95000}{\left(\frac{19t}{10}+9\right)^2}\\$Simplify/rewrite to obtain:$\\\\D'(t)=-\dfrac{9500000}{\left(19t+90\right)^2}[/tex]
Therefore, when t=115 days
[tex]D'(115)=-\dfrac{9500000}{\left(19(115)+90\right)^2}\\D'(115)=-1.8355[/tex]
