The price-demand and cost functions for the production of microwaves are given as
p=235−x40
and
C(x)=14000+70x,
where x is the number of microwaves that can be sold at a price of p dollars per unit and C(x) is the total cost (in dollars) of producing x units.
(A) Find the marginal cost as a function of x.
C′(x) =
(B) Find the revenue function in terms of x.
R(x) =
(C) Find the marginal revenue function in terms of x.
R′(x) =
(D) Evaluate the marginal revenue function at x=1500.
R′(1500) =
(E) Find the profit function in terms of x.
P(x) =
(F) Evaluate the marginal profit function at x=1500.
P′(1500) =
If you can follow the one I did, you can do the rest.
(A) The marginal cost, denoted as C'(x), represents the rate of change of the total cost function with respect to the quantity produced.
To find C'(x), we need to take the derivative of the cost function, C(x), with respect to x.
C(x) = 14000 + 70x
Taking the derivative of C(x) with respect to x:
C'(x) = d/dx(14000 + 70x)
= 0 + 70
= 70
Therefore, the marginal cost function is C'(x) = 70.
(B) The revenue function, denoted as R(x), represents the total revenue earned from selling x units of the product.
To find R(x), we need to multiply the number of units sold, x, by the price per unit, p.
Given the price-demand function:
p = 235 - 40x
R(x) = xp
Substituting the price function into the revenue function:
R(x) = x(235 - 40x)
= 235x - 40x^2
Therefore, the revenue function in terms of x is R(x) = 235x - 40x^2.
(C) The marginal revenue function, denoted as R'(x), represents the rate of change of the revenue function with respect to the quantity produced.
To find R'(x), we need to take the derivative of the revenue function, R(x), with respect to x.
R(x) = 235x - 40x^2
Taking the derivative of R(x) with respect to x:
R'(x) = d/dx(235x - 40x^2)
= 235 - 80x
Therefore, the marginal revenue function is R'(x) = 235 - 80x.
(D) To evaluate the marginal revenue function at x = 1500, we substitute x = 1500 into the marginal revenue function R'(x).
R'(1500) = 235 - 80(1500)
= 235 - 120000
= -119765
Therefore, R'(1500) = -119765.
(E) The profit function, denoted as P(x), represents the difference between total revenue and total cost.
To find P(x), we subtract the cost function C(x) from the revenue function R(x).
P(x) = R(x) - C(x)
= (235x - 40x^2) - (14000 + 70x)
= 235x - 40x^2 - 14000 - 70x
= -40x^2 + 165x - 14000
Therefore, the profit function in terms of x is P(x) = -40x^2 + 165x - 14000.
(F) To evaluate the marginal profit function at x = 1500, we need to find the derivative of the profit function, P(x), with respect to x and then substitute x = 1500.
To find P'(x), we take the derivative of the profit function:
P(x) = -40x^2 + 165x - 14000
P'(x) = d/dx(-40x^2 + 165x - 14000)
= -80x + 165
Substituting x = 1500 into P'(x):
P'(1500) = -80(1500) + 165
= -120000 + 165
= -119835
Therefore, P'(1500) = -119835.