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Raggs, Ltd. a clothing​ firm, determines that in order to sell x​ suits, the price per suit must be p=120 - 0.5 x. It also determines that the total cost of producing x suits is given by C (x)=2500 + 0.25 x ^2. ​a) Find the total​ revenue, R (x). ​b) Find the total​ profit, P (x). ​c) How many suits must the company produce and sell in order to maximize​ profit? ​d) What is the maximum​ profit? ​e) What price per suit must be charged in order to maximize​ profit?

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Answer:

A) R(x) = 120x - 0.5x^2

B) P(x) = - 0.75x^2 + 120x - 2500

C) 80

D) 2300

E) 80

Explanation:

Given the following :

Price of suit 'x' :

p = 120 - 0.5x

Cost of producing 'x' suits :

C(x)=2500 + 0.25 x^2

A) calculate total revenue 'R(x)'

Total Revenue = price × total quantity sold, If total quantity sold = 'x'

R(x) = (120 - 0.5x) * x

R(x) = 120x - 0.5x^2

B) Total profit, 'p(x)'

Profit = Total revenue - Cost of production

P(x) = R(x) - C(x)

P(x) = (120x - 0.5x^2) - (2500 + 0.25x^2)

P(x) = 120x - 0.5x^2 - 2500 - 0.25x^2

P(x) = - 0.5x^2 - 0.25x^2 + 120x - 2500

P(x) = - 0.75x^2 + 120x - 2500

C) To maximize profit

Find the marginal profit 'p' (x)'

First derivative of p(x)

d/dx (p(x)) = - 2(0.75)x + 120

P'(x) = - 1.5x + 120

-1.5x + 120 = 0

-1.5x = - 120

x = 120 / 1.5

x = 80

D) maximum profit

P(x) = - 0.75x^2 + 120x - 2500

P(80) = - 0.75(80)^2 + 120(80) - 2500

= -0.75(6400) + 9600 - 2500

= -4800 + 9600 - 2500

= 2300

E) price per suit in other to maximize profit

P = 120 - 0.5x

P = 120 - 0.5(80)

P = 120 - 40

P = $80

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