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?

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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