Solve the problem.
-The price $p$ and the quantity $x$ sold of a certain product obey the demand equation
$$p = - \frac { 1 } { 6 } x + 200 , \quad 0 \leq x \leq 1,200 \text {. }$$
What quantity $x$ maximizes revenue? What is the maximum revenue?
A) $600 ; \$ 60,000$
B) $900 ; \$ 45,000$
C) $300 ; \$ 45,000$
D) 1,$200 ; \$ 60,000$

Question

Quizplus · Accepted Answer

Revenue $R$ is given by the product of price $p$ and quantity $x$, so $R = px$. Substituting the given demand equation into this formula gives $R = (-\frac{1}{6}x + 200)x = -\frac{1}{6}x^2 + 200x$. To maximize revenue, we take the derivative of $R$ with respect to $x$ and set it to zero: $\frac{dR}{dx} = -\frac{1}{3}x + 200 = 0$, solving for $x$ gives $x = 600$. Substituting $x = 600$ back into the revenue equation gives the maximum revenue: $R = -\frac{1}{6}(600)^2 + 200(600) = 60,000$.

Solve the Problem ppp And the Quantity xxx Sold of a Certain Product Obey the Demand Equation

Solve the Problem $p$ And the Quantity $x$ Sold of a Certain Product Obey the Demand Equation