a film company is deciding on the price of the video release on one of its films. its marketing people…

Explanation:

Step1: Define the profit function

The revenue $R$ is the product of the price $p$ and the quantity $q$, so $R = pq$. Given $q=200000 - 10000p$, then $R=p(200000 - 10000p)=200000p-10000p^{2}$. The cost $C$ of producing $q$ copies is $C = 3q$. Substituting $q = 200000 - 10000p$ into the cost - function, we get $C=3(200000 - 10000p)=600000-30000p$. The profit function $P$ is $P=R - C$. So $P=(200000p-10000p^{2})-(600000 - 30000p)=- 10000p^{2}+230000p - 600000$.

Step2: Find the vertex of the quadratic function

For a quadratic function $y = ax^{2}+bx + c$ ($a\neq0$), the $x$ - coordinate of the vertex (which gives the maximum value when $a\lt0$) is $x=-\frac{b}{2a}$. In our profit function $P=-10000p^{2}+230000p - 600000$, where $a=-10000$, $b = 230000$, and $c=-600000$. Using the formula $p=-\frac{b}{2a}$, we have $p=-\frac{230000}{2\times(-10000)}=\frac{230000}{20000}=11.5$.

Answer:

$11.5$