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 $R = p \times x$. Substituting $x = -6p + 120$ into the revenue equation gives $R = p(-6p + 120)$. Expanding this, we get $R = -6p^2 + 120p$. To find the maximum revenue, we take the derivative of $R$ with respect to $p$ and set it to zero. The derivative is $\frac{dR}{dp} = -12p + 120$. Setting this equal to zero gives $p = 10$. Substituting $p = 10$ back into the demand equation gives $x = -6(10) + 120 = 60$. The maximum revenue is then $R = 10 \times 60 = 600$.

Solve the Problem x=−6p+120,0≤p≤20x = - 6 p + 120 , \quad 0 \leq p \leq 20x=−6p+120,0≤p≤20

Solve the Problem $x = - 6 p + 120 , \quad 0 \leq p \leq 20$