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
Answer: D. Both thought processes are linear
Explanation: I think I'm not too certain.
Important dsiclamer: there was a type in the question you enter 26,000 while in the textbook is for 20,000
Answer:
a. Decrease $1,200,000
Explanation:
Income before internal transfer:
revenue 3150
cost 1050
gross 2100
fixed (2100)
operating 0
external engine purchase (3000)
net (3000)
After internal change:
revenue 1050
cost (960)
gross profit 90
fixed (2100)
operating (2010)
internal engine purchase (1,050)
net (3,060)
difference -3060--3000 = 60
20,000 units x 60 = 1,200,000
Explanation:
this is easy to understand and even to answer so if I tell u the answer
how will u learn to do things yourself. By me being a business Woman i learn't that in order to gain sumthing in life u have to do it yourself
Going to a community and then transferring
