Solve the problem.
-The manufacturer of a CD player has found that the revenue $\mathrm { R }$ (in dollars) is $\mathrm { R } ( \mathrm { p } ) = - 5 \mathrm { p } ^ { 2 } + 1310 \mathrm { p }$, when the unit price is $p$ dollars. If the manufacturer sets the price $p$ to maximize revenue, what is the maximum revenue to the nearest whole dollar?
A) $\$ 85,805$
B) $\$ 171,610$
C) $\$ 343,220$
D) $\$ 686,440$

Question

Quizplus · Accepted Answer

To find the maximum revenue, we first need to find the vertex of the parabola represented by the revenue function $R(p) = -5p^2 + 1310p$. The vertex form of a parabola is $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex. The x-coordinate of the vertex, which gives the price $p$ that maximizes revenue, can be found using the formula $h = -\frac{b}{2a}$ for a quadratic equation in the form of $ax^2 + bx + c$. Here, $a = -5$ and $b = 1310$, so $p = -\frac{1310}{2(-5)} = 131$. Substituting $p = 131$ back into the revenue function gives the maximum revenue: $R(131) = -5(131)^2 + 1310(131) = 85,805$.

Solve the Problem R\mathrm { R }R (In Dollars) Is R(p)=−5p2+1310p\mathrm { R } ( \mathrm { p } ) = - 5 \mathrm { p } ^ { 2 } + 1310 \mathrm { p }R(p)=−5p2+1310p

Solve the Problem $\mathrm { R }$ (In Dollars) Is $\mathrm { R } ( \mathrm { p } ) = - 5 \mathrm { p } ^ { 2 } + 1310 \mathrm { p }$