The unit selling price p (in dollars) and the quantity demanded x (in pairs) of a certain brand of women’s shoes are given by the demand equation
p(x) = 100e^-0.0001x f or 0 _< x _< 20,000
a. Find the revenue function,R. (Hint: R(x)= x(p(x)), since the
revenue function is the unit selling price at a demand level of x units
times the number of units demanded.)
b. Find the marginal revenue function, R'.
c. What is the marginal revenue when, x = 10 ?
a. To find the revenue function, R(x), we use the formula R(x) = x(p(x)), where x is the quantity demanded and p(x) is the unit selling price. In this case, the unit selling price is given by the demand equation p(x) = 100e^(-0.0001x).
Substituting p(x) into the formula, we have:
R(x) = x(100e^(-0.0001x))
b. To find the marginal revenue function, R', we differentiate the revenue function, R(x), with respect to x. The derivative of R(x) can be found using the product rule.
Applying the product rule to R(x) = x(100e^(-0.0001x)), we get:
R'(x) = 100e^(-0.0001x) - x * (0.0001) * 100e^(-0.0001x)
R'(x) = 100e^(-0.0001x) - 0.01x * e^(-0.0001x)
c. To find the marginal revenue when x = 10, we substitute x = 10 into the marginal revenue function, R'(x), derived in part b.
R'(10) = 100e^(-0.0001 * 10) - 0.01 * 10 * e^(-0.0001 * 10)
Now we can calculate the value of R'(10) using a calculator or computational software.