Question

A cosmetics company is planning the introduction and promotion of a new lipstick line. The marketing research department has found that the demand in a particular city is given approximately by p=6x, 0≤x≤2, where x thousand lipsticks were sold per week at a price of p dollars each. At what price will the weekly revenue be maximized?

Answer 1

The weekly revenue for a new lipstick line will be maximized when the product is priced at $6 per lipstick, given the demand function p = 6x, where x is the quantity sold in thousands per week.

Step-by-step explanation:

To determine the price at which weekly revenue for the new lipstick line will be maximized, we need to understand the concept of price elasticity of demand and how it relates to total revenue. In this case, the demand function is given by p = 6x, where p is the price in dollars and x represents the thousands of lipsticks sold per week. Total revenue (R) is the product of price (p) and quantity (x), so R = p × x or R = 6x^2.

To maximize revenue, we need to find the value of x that maximizes R. Differentiating R with respect to x and setting the derivative equal to zero gives us the quantity x that maximizes revenue. Doing this, we find that x = 1 thousand lipsticks per week maximizes the revenue. Substituting this back into the demand function, we find that the price p which maximizes revenue is $6 per lipstick.