Question

Suppose that the demand function for a product is given by ​D(p)equals=StartFraction 50 comma 000 Over p EndFraction 50,000 p and that the price p is a function of time given by pequals=1.91.9tplus+99​, where t is in days. ​a) Find the demand as a function of time t. ​b) Find the rate of change of the quantity demanded when tequals=115115 days. ​a)​ D(t)equals=nothing ​(Simplify your​ answer.)

Answer 1

Answer:

(a)$D(t)=\dfrac{50000}{1.9t+9}$

(b)$D'(115)=-1.8355$

Step-by-step explanation:

The demand function for a product is given by :

$D(p)=\dfrac{50000}{p}$

Price, p is a function of time given by $p=1.9t+9$, where t is in days.

(a)We want to find the demand as a function of time t.

$\text{If } D(p)=\dfrac{50000}{p}, and p=1.9t+9\\Then:\\D(t)=\dfrac{50000}{1.9t+9}$

(b)Rate of change of the quantity demanded when t=115 days.

$\text{If } D(t)=\dfrac{50000}{1.9t+9}$

$\dfrac{\mathrm{d}}{\mathrm{d}t}\left[\dfrac{50000}{\frac{19t}{10}+9}\right]}}=50000\cdot \dfrac{\mathrm{d}}{\mathrm{d}t}\left[\dfrac{1}{\frac{19t}{10}+9}\right]}$

$=-50000\cdot\dfrac{d}{dt} \dfrac{\left[\frac{19t}{10}+9\right]}{\left(\frac{19t}{10}+9\right)^2}}}$

$=\dfrac{-50000(1.9\frac{d}{dt}t+\frac{d}{dt}9)}{\left(\frac{19t}{10}+9\right)^2}}}$

$=-\dfrac{95000}{\left(\frac{19t}{10}+9\right)^2}\\ Simplify/rewrite to obtain: \\\\D'(t)=-\dfrac{9500000}{\left(19t+90\right)^2}$

Therefore, when t=115 days

$D'(115)=-\dfrac{9500000}{\left(19(115)+90\right)^2}\\D'(115)=-1.8355$