Answer :
Answer:
The ticket price that maximizes revenue is $18.10
Explanation:
Hi, first we need to construct the revenue equation in terms of the additional dollar charge (that would be X). That is:
[tex]Revenue=Price*Quantity[/tex]
[tex]Revenue=(8.5-X)(1,800-65X)[/tex]
So we expand it:
[tex]Revenue=15,300-552.5X+1,800X-65X^{2}[/tex]
[tex]Revenue=-65X^{2} +1247.5X+15,300[/tex]
This is a parabola, and we need to find its vertex, which in our case that would be the maximum additional dollar charge in order to obtain the highest revenue possible, to find the vertex, we need to consider that:
[tex]Y(X)=AX^{2}+ BX+C[/tex]
And to find the X-coordenate we have to use the following equation.
[tex]Vertex(X)=\frac{-B}{2A}[/tex]
In our case, A= -65; B= 1,247.5, so, all should look like this:
[tex]Vertex (X)=\frac{-(1247.50}{2(-65)} =9.6[/tex]
That means, we need to make 9.6 increments of $1 in order to obtain the max revenue possible, therefore, the price would be
Price = $8.50 + $1(9.6)= $8.50 + $9.6 =$18.10
Best of luck.