we have the linear equation
[tex]p=(-\frac{1}{5})x+200[/tex]
Part A
the Revenue is equal to
R(x)=p*x
substitute
[tex]R(x)=\lbrack(-\frac{1}{5})x+200\rbrack\cdot x[/tex][tex]R(x)=-\frac{1}{5}x^2+200x[/tex]
Part B
the function of revenue is a quadratic equation (vertical parabola opens downward)
the vertex represents a maximum
using a graphing tool
see the attached figure
the vertex is the point (500, 50000)
therefore
the value of x that maximizes the revenue is
x=500
Part C
For x=500
Find out the unit price
[tex]\begin{gathered} p=(-\frac{1}{5})(500)+200 \\ p=100 \end{gathered}[/tex]
the unit price is p=$100
Part D
the maximum revenue is the y-coordinate of the vertex
the vertex is (500, 50,000)
therefore
the maximum revenue is $50,000
