SOLUTION: Find the max revenue of {{{R=-15p^2+300p+1200}}} by use of a graph.

Algebra ->  Graphs -> SOLUTION: Find the max revenue of {{{R=-15p^2+300p+1200}}} by use of a graph.      Log On


   

Question 147714: Find the max revenue of R=-15p%5E2%2B300p%2B1200 by use of a graph.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
If we graph R=-15p%5E2%2B300p%2B1200 using a graphing calculator, we get:

Graph of R=-15p%5E2%2B300p%2B1200

From the graph, we can see that the highest point on the graph is (10,2700) (you can use the "min/max" feature to find this point).

Since the highest point has the y-value 2700, this means that the max revenue is $2700. This max revenue occurs when the price is $10 (since the x-value of the vertex is x=10).
Also, the reason why the graph is parabolic is because the equation is a quadratic.