Solution
Step 1:
Write the revenue function
[tex]\text{r = -3p}^2\text{ + 60p + 1060}[/tex]Step 2:
The first derivative of the revenue should be used to determine the price to maximize sales.
[tex]\begin{gathered} \text{r = -3p}^2\text{ + 60p + 1060} \\ \frac{dr}{dp}\text{ = -6p + 60} \\ -6p\text{ + 60 = 0} \\ 6p\text{ = 60} \\ p\text{ = }\frac{60}{6} \\ p\text{ = 10} \end{gathered}[/tex]