Question

Revenue The demand equation for one type of computer networking system is

p=D(q)=500 q e⁻⁰.⁰⁰¹⁶ q²
where p is the price (in dollars) and q is the quantity of servers sold per month. Find the values of q and p that maximize revenue.

Answer 1

To maximize revenue, we need to find the values of q and p that maximize the function R(q) = p * q, where p is the price and q is the quantity of servers sold per month. In this case, we are given the demand equation as p = D(q) = 500qe^(-0.0016q^2). To find the maximum revenue, we need to find the critical points of the revenue function and determine which one corresponds to a maximum.

Step-by-step explanation:

To maximize revenue, we need to find the values of q and p that maximize the function R(q) = p * q, where p is the price and q is the quantity of servers sold per month. In this case, we are given the demand equation as p = D(q) = 500qe^(-0.0016q^2). To find the maximum revenue, we need to find the critical points of the revenue function and determine which one corresponds to a maximum. We can do this by finding the derivative of R(q) with respect to q and setting it equal to zero, then solving for q. Once we have the value of q, we can substitute it back into the demand equation to find the corresponding value of p.

Step-by-step solution:

1. Find the derivative of R(q) = p * q with respect to q: R'(q) = p + q * dp/dq.
2. Substitute the demand equation p = D(q) into the derivative: R'(q) = D(q) + q * d(D(q))/dq
3. Set R'(q) = 0 and solve for q to find the critical points.
4. After finding the value of q that maximizes revenue, substitute it back into the demand equation p = D(q) to find the corresponding value of p.