A community theater uses the function
p(d) = -4d^2 + 2000d – 100
to model the profit (in dollars) expected in a weekend when the tickets to a comedy show
are priced at d
dollars each.
a. Write and solve an equation to find out the prices at which the theater would earn
$1,500 in profit from the comedy show each weekend. Show your reasoning.
b. At what price would the theater make the maximum profit, and what is that maximum profit? Show your reasoning.
Part a)
For p(d)=$1,500
substitute in the given equation
1500=-4d^2 + 2000d – 100
solve for d
-4d^2+2000d-100-1500=0
-4d^2+2000d-1600=0
Using the formula to solve quadratic equation
we have
a=-4
b=2000
c=-1600
substitute
![d=\frac{-2000\pm\sqrt[]{(2000^2)-4(-4)(-1600)}}{2(-4)}](https://img.qammunity.org/2023/formulas/mathematics/college/3k06noxr3zxvpya57zy9ydxm6g40f8ssmr.png)
![\begin{gathered} d=\frac{-2000\pm\sqrt[]{3,974,400}}{-8} \\ \\ d=(-2000\pm1,993.59)/(-8) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lyouiyqbwiuogfb5l6pt6g6inge7vgsr3a.png)
the solutions for d are
d=$0.8
d=$499
Part b) At what price would the theater make the maximum profit, and what is that maximum profit? Show your reasoning.
we know that
The maximum profit will be equal to the y-coordinate of the vertex of the quadratic equation
so
p(d) = -4d^2 + 2000d – 100
Find the vertex
using a graphing tool
see the attached figure
the vertex is the point (250, 249,900)
that means
the maximum profit is $249,900 and the price that would the theater make the maximum profit is $250