The Solution:
Given the profit function below:
[tex]P=-0.02n^2+3.40n-16[/tex]
To maximize profit, we have to differentiate the given function P with respect to n, and then equate to zero.
[tex]\begin{gathered} \frac{dP}{dx}=0 \\ \\ \frac{dP}{dx}=2(-0.02)n+3.40=0 \\ \\ -0.04n+3.40=0 \end{gathered}[/tex]
Solving for n, we get
[tex]\begin{gathered} -0.04n=-3.40 \\ \text{ Dividing both sides by -0.04} \\ n=\frac{-3.40}{-0.04}=85\text{ thousand units of DVD} \end{gathered}[/tex]
Thus, the correct answer to part (a) is 85 thousand units of DVD.
Part (b)
We are required to find the maximum profit.
We shall substitute 85 for n in the given profit function.
[tex]\begin{gathered} P(n)=-0.02n^2+3.40n-16 \\ \\ P(85)=-0.02(85)^2+3.40(85)-16 \end{gathered}[/tex][tex]\begin{gathered} P(85)=128.5 \\ So, \\ The\text{ maximum profit is }128.5\text{ thousand of dollars.} \end{gathered}[/tex]
Therefore, the maximum profit is 128.5 thousand of dollars.
