Question

The demand, q, for a particular product is expressed as a function of the price, p, by the formula,q= 320 - 0.15pa) Express the total revenue, R, as a function of the price, p. R(p) = P*QR(p)= ____b) Find the price, p, which maximizes revenue.P= $____c) What is the maximum revenue at that production level?Revenue= $____

Answer 1

Answer:

a) R(p) = 320p - 0.15p²

b) P = $1066.67

c) Revenue = $170666.67

Explanation:

We know that the demand Q is given by

Q = 320 - 0.15p

Then, the revenue R is equal to P*Q, so by replacing the expression above, we get

R(p) = p*Q

R(p) = p(320 - 0.15p)

R(p) = 320p - 0.15p²

Now, to find the price that maximizes the revenue, we need to find the coordinate of the vertex of a parabola. This coordinate is equal to -b/2a, where b is the number besides p and a is the number beside p². So, the price that maximizes revenue is

$\begin{gathered} p=\frac{-b}{2a} \\ \\ p=\frac{-320}{2(-0.15)}=1066.67 \end{gathered}$

Finally, to know the maximum revenue, we need to replace p = 1066.67 on the equation for R(p).

$\begin{gathered} R(p)=320p-0.15p^2 \\ R(p)=320(1066.67)-0.15(1066.67)^2 \\ R(p)=170666.67 \end{gathered}$

Therefore, the answers are

a) R(p) = 320p - 0.15p²

b) P = $1066.67

c) Revenue = $170666.67