The revenue will be 12,000 when the price of the product is $100
Given :
The monthly revenue of a certain company
[tex]R=820p-7p^2[/tex]
We need to find out p when Revenue R is 12000
Lets replace R with 12000 and solve for p
[tex]12000=820p-7p^2\\820p-7p^2=12000\\-7p^2+820p-12000=0[/tex]
Apply quadratic formula
[tex]p=\frac{-820\pm \sqrt{820^2-4\left(-7\right)\left(-12000\right)}}{2\left(-7\right)}\\p=\frac{-820\pm \:580}{2\left(-7\right)}\\p=\frac{-820+580}{2\left(-7\right)},\:p=\frac{-820-580}{2\left(-7\right)}\\p=\frac{120}{7},\:p=100[/tex]
Given that price must be greater than 50
So the revenue will be 12000 when the price of the product is $100
Learn more : brainly.com/question/21503855
