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

(a)
`D(t)=(50000)/(1.9t+9)`

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

Explanation:

The demand function for a product is given by :

`D(p)=(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)=(50000)/(p),$ and p=1.9t+9\\Then:\\D(t)=(50000)/(1.9t+9)`

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

`\text{If } D(t)=(50000)/(1.9t+9)`

`\frac{\mathrm{d}}{\mathrm{d}t}\left[(50000)/((19t)/(10)+9)\right]}}=50000\cdot \frac{\mathrm{d}}{\mathrm{d}t}\left[(1)/((19t)/(10)+9)\right]}`

`=-50000\cdot(d)/(dt) (\left[(19t)/(10)+9\right])/(\left((19t)/(10)+9\right)^2)}}`

`=(-50000(1.9(d)/(dt)t+(d)/(dt)9))/(\left((19t)/(10)+9\right)^2)}}`

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

Therefore, when t=115 days

`D'(115)=-(9500000)/(\left(19(115)+90\right)^2)\\D'(115)=-1.8355`