Solve the problem.
-The price $p$ (in dollars) and the quantity $x$ sold of a certain product obey the demand equation $x = - 6 p + 120 , \quad 0 \leq p \leq 20$.
What quantity $x$ maximizes revenue? What is the maximum revenue?
A) $60 ; \$ 600$
B) 90; $\$ 450$
C) $30 ; \$ 450$
D) $120 ; \$ 600$

Question

Quizplus · Accepted Answer

Revenue $R$ is given by $R = xp = (-6p + 120)p$. To maximize revenue, differentiate $R$ with respect to $p$ and set the derivative equal to zero. The derivative $R' = -12p + 120$. Setting $R' = 0$ gives $p = 10$. Substituting $p = 10$ into the demand equation gives $x = -6(10) + 120 = 60$. The maximum revenue is $R = 10 \times 60 = \$600$.

Solve the Problem ppp (In Dollars) and the Quantity xxx Sold of a Certain Product Obey the Demand Equation

Solve the Problem $p$ (In Dollars) and the Quantity $x$ Sold of a Certain Product Obey the Demand Equation