## anonymous 5 years ago 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

1. anonymous

\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

2. anonymous

I guess d. is the answer

3. anonymous

thanks

4. anonymous

Thank Wolfram Research.