Answer to Question #113897 in Calculus for Elizabeth

Question

A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the number of cakes sold per day, and p is price. The total cost function of the company is given by c = (30+5x) 2 where x is previously defined, and c is total cost. A. Find the revenue and marginal revenue functions [Hint: revenue is price multiplied by quantity i.e. revenue = price × quantity] .B. Find the fixed cost and marginal cost function [Hint: fixed cost does not change with quantity produced] . C. Find the profit function [Hint: profit is revenue minus total cost] . D. Find the quantity that maximizes profit.

Answer 1

A) Find the revenue and marginal revenue functions

Revenue function

R = PX

R = (580-10x) x

R = 580x - 10x²

Marginal revenue is the derivative of revenue function:

R = 580x - 10x²

R' = (1)(580)x^1-1 - (2)(10)x^2-1

R' = 580 - 20x

B) Find the fixed cost and marginal cost function

C = (30 + 5x)2

C = 60 + 10x

Fixed costs are costs not dependent on the quantity. Thus, fixed costs = 60, which is the constant number in the cost equation above.

Marginal cost is the derivative of the cost function.

C = 60 + 10x

C' = 0 + (1)10x^1-1 -------> note: Derivative of constant is zero. So in this case, the derivative of 60 is zero

C' = 10

C) Find the profit function

Profit = Revenue - Total Cost

P = (580x - 10x²) - (60 + 10x)

P = 580x - 10x² - 60 - 10x

P = -10x² + 570x - 60

D. Find the quantity that maximizes profit

To get the maximum quantity (x) derive the profit function and equate it to zero.

P = -10x² + 570x - 60

P' = -(2)(10)x^2-1 + (1)(570)x^1-1 - 0 -------> (note: Derivative of constant is zero in this case derivative of 60 is 0)

P' = -20x + 570 -------> ( P' = 0 to get maximum value of x)

0 = -20x + 570

20x = 570

x = 28.5

x ≈ 29 cakes