Question

Suppose the sales from a product generates a revenue (R)
r=-7.3p^2+320p


where p is the price of the product in dollars.

Find the prices of the product that will generate a revenue greater than $ 3000.

a. Price > $ 14.00 and price < $ 32
b. Price > $ 16 and price < $ 50
c. Price > $13.58 and price < $ 30.25
d. Price > $ 13.28 and price < $ 30.20
e. Price < $ 13.50 and Price > $ 30. 56

Answer 1

\text{Reduce}\left[3000<320 p-\frac{73}{10} p^2\right]= \[\text{Reduce}\left[3000<320 p-\frac{73}{10} p^2\right]=\frac{100}{73} \left(16-\sqrt{37}\right)<p<\frac{100}{73} \left(16+\sqrt{37}\right) \]
= 13.5853 < p < 30.2504

Answer 2

I guess d. is the answer

Answer 3

thanks

Answer 4

Thank Wolfram Research.