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vazorg [7]
3 years ago
13

Revenue

Mathematics
1 answer:
denis23 [38]3 years ago
7 0

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

p=ce^{kx}

Substitute  the given values for p and x to get two equations in c and k

40 = ce^{1200k} \\45 = ce^{1000k}

Dividing on by other we get

\frac{45}{40} =e^{-200k} \\-200 k = ln (45/40) = 0.117583\\k = -0.000589

Substitute value of k in any one equation

45 = ce^{-0.589} \\c=45.02651

b) Revenue of the product is demand and price

i.e. R(x) = p*x = 45.02651xe^{-0.000589x}

Use Calculus derivative test to find max Revenue

R'(x) = 45.02651 e^{-0.000589x}-45.02651*0.000589 x e^{-0.000589x}\\

EquateI derivative to 0

1-0.000589x =0

x = 1698.037

When x = 1698 and p = 16.56469

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