Answer:
Step-by-step explanation:
Given that a small business assumes that the demand function for one of its new products can be modeled by
[tex]p=ce^{kx}[/tex]
Substitute the given values for p and x to get two equations in c and k
[tex]40 = ce^{1200k} \\45 = ce^{1000k}[/tex]
Dividing on by other we get
[tex]\frac{45}{40} =e^{-200k} \\-200 k = ln (45/40) = 0.117583\\k = -0.000589[/tex]
Substitute value of k in any one equation
[tex]45 = ce^{-0.589} \\c=45.02651[/tex]
b) Revenue of the product is demand and price
i.e. R(x) = p*x = [tex]45.02651xe^{-0.000589x}[/tex]
Use Calculus derivative test to find max Revenue
R'(x) = [tex]45.02651 e^{-0.000589x}-45.02651*0.000589 x e^{-0.000589x}\\[/tex]
EquateI derivative to 0
1-0.000589x =0
x = 1698.037
When x = 1698 and p = 16.56469
