Notice that if we expand the product for the function R(x), we obtain a quadratic equation:
[tex]R(x)=(500+125x)(50-5x)[/tex]
The graph of a quadratic function is a parabola, and the x-coordinate of the vertex of the parabola is located between the zeros of the function.
The zeros of the function are given by the equations:
[tex]\begin{gathered} 500+125x=0 \\ \Rightarrow x_1=\frac{-500}{125}=-4 \\ \\ 50-5x=0 \\ \Rightarrow x_2=\frac{50}{5}=10 \end{gathered}[/tex]
Then, the x-coordinate of the vertex of the parabola is:
[tex]x_v=\frac{x_1+x_2}{2}=\frac{-4+10}{2}=\frac{6}{2}=3[/tex]
For x=3, the price of the yearbooks is:
[tex]50-5(3)=35[/tex]
The possible maximum revenue would be:
[tex]\begin{gathered} R(3)=(500+125\cdot3)(50-5\cdot3) \\ =(875)(35) \\ =30,625 \end{gathered}[/tex]
And the amount of yearbooks they would sell is:
[tex]500+125\cdot3=875[/tex]
Therefore, the answers are:
Price of the yearbooks: $35
Maximum possible revenue: $30,625
Amount of yearbooks they will sell: 875
