Answer:
The maximum revenue is $1,20,125 that occurs when the unit price is $155.
Step-by-step explanation:
The revenue function is given as:
[tex]R(p) = -5p^2 + 1550p[/tex]
where p is unit price in dollars.
First, we differentiate R(p) with respect to p, to get,
[tex]\dfrac{d(R(p))}{dp} = \dfrac{d(-5p^2 + 1550p)}{dp} = -10p + 1550[/tex]
Equating the first derivative to zero, we get,
[tex]\dfrac{d(R(p))}{dp} = 0\\\\-10p + 1550 = 0\\\\p = \dfrac{-1550}{-10} = 155[/tex]
Again differentiation R(p), with respect to p, we get,
[tex]\dfrac{d^2(R(p))}{dp^2} = -10[/tex]
At p = 155
[tex]\dfrac{d^2(R(p))}{dp^2} < 0[/tex]
Thus by double derivative test, maxima occurs at p = 155 for R(p).
Thus, maximum revenue occurs when p = $155.
Maximum revenue
[tex]R(155) = -5(155)^2 + 1550(155) = 120125[/tex]
Thus, maximum revenue is $120125 that occurs when the unit price is $155.
