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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

Respuesta :

temdan2001

Answer:

A.) R = 580x - 10x^2

Marginal revenue = 580−20x

B.) Fixed cost = 60

Marginal cost = 10

C.) P = -10x^2 + 570x - 60

D.) X = 29

Step-by-step explanation:

A.) The revenue function

R= P × quantity x

R = (580-10x)x

R = 580x - 10x^2

The marginal revenue function is differential of R

dR/dx = (580x-10x^2)'

dR/dx = 580−20x

B.) The fixed cost

C = (30 + 5x)2

C = 60 + 10x

fixed cost does not change with quantity X produced.That is, fixed costs doesn't dependent on the quantity. It is the constant number in the cost equation above . Therefore

fixed costs = 60.

The marginal cost function

Marginal cost is the derivative of the cost function. That is

C = 60 + 10x

dC/dx = 10

C.) The profit function

Profit = Revenue - Total Cost

P = (580x - 10x^2) - (60 + 10x)

P = 580x - 10x^2 - 60 - 10x

P = -10x^2 + 570x - 60

D.) The quantity that maximizes profit

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

P = -10x^2 + 570x - 60

dP/dx = -(2)(10)x + (1)(570) - 0

dP/dx = -20x + 570

To get maximum value of x, dP/dx= 0

0 = -20x + 570

20x = 570

x = 28.5

x = 29 cakes approximately

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