The CarryItAll mini-van, a popular vehicle among soccer moms, obeys the demand e
ID: 3007252 • Letter: T
Question
The CarryItAll mini-van, a popular vehicle among soccer moms, obeys the demand equation p =/50 x + 16,000. The cost of producing x vans is given by the function C(x) = 13040x + 20,000. Express the revenue R as a function of x. Express the profit P as a function of x. Find the value of x that maximizes profit. What is the maximum profit? What price should be charged in order to maximize profit? R(x) = (Simplify your answer. Do not factor.) P(x) = (Simplify your answer. Do not factor.) What quantity will maximize the profit? vans What is the maximum profit? $ million (Round to the nearest million dollars as needed.) What price should be charged for the maximum profit? $ (Round to the nearest dollar as needed.)Explanation / Answer
here the cost function is given as
C(x) = 13040x + 20000
now the revenue function is given as
R(x) = number of units*unit price
number of units = the demand function = P = -x/50 + 16000
=> R(x) = [-x/50 + 16000]*x
R(x) = -x^2/50 + 16000x
b>
the profit function = R(x) - C(x)
= [-x^2/50 + 16000x] - [13040x + 20000]
P(x) = -x^2/50 + 2960x - 20000
c> for maximixing the profit
th profit function is parabolic in nature
P(x) = -x^2/50 + 2960x - 20000
so the maximum profit will be at the vertex of the parabola
the x coordinate of the vertex is found as
x = -b/2a
b = 2960 and a = -1/50
=> x = -2960*-25 = 74000
therefore x = 74000 vans will maximize the profit
moximum profit is = P(x=74000) = -(74000)^2/50 + 2960(74000) - 20000 = 109500000 = 109.5 = 110 million dollars
d>
for maximum profit
C(x=74000) = 13040(74000) + 20000 = 964980000
now price that need to be chared = cost + profit = 964980000 + 109500000 = 1074480000 = 1074.48 = 1075 million dollars