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
it's important to invest so you can have a better life once that thing you invested in makes you money and not all of them do so keep that in mind
The total amount of money being transferred into and out of a business
