Maximum profit based on revenue and costs using Calculus?
30.September, 2009
A study done for management shows that the revenue, R, in dollars, as a function of number, x, of units produced can be modeled by R(x) = 12x – 0.001×2 while the cost, C, in dollars, can be modeled by C(x) = 5000 + 2x. What will the maximum profit be, in dollars?
My initial impulse is to plug C(x) in for all x values in R(x), get the derivative of R(x) and then set it equal to 0. I’m not sure if this is correct or not though.
Profit is equal to revenue minus cost. Define profit function P(x):
P(x) = R(x) – C(x)
P(x) = 12x – 0.001x^2 – 5000 – 2x
P(x) = -0.001x^2 + 10x – 5000
Differentiate the profit function.
P’(x) = -0.002x + 10
Set equal to zero.
0 = -0.002x + 10
0.002x = 10
x = 5000
Maximum profit occurs with 5000 units sold. What is the value of P(5000)?
P(5000) = -0.001(5000)^2 + 10*5000 – 5000
P(5000) = 20,000
Twenty thousand dollars is the maximum profit. Good luck. =)
01.October, 2009 um 4:31 am
Profit is equal to revenue minus cost. Define profit function P(x):
P(x) = R(x) – C(x)
P(x) = 12x – 0.001x^2 – 5000 – 2x
P(x) = -0.001x^2 + 10x – 5000
Differentiate the profit function.
P’(x) = -0.002x + 10
Set equal to zero.
0 = -0.002x + 10
0.002x = 10
x = 5000
Maximum profit occurs with 5000 units sold. What is the value of P(5000)?
P(5000) = -0.001(5000)^2 + 10*5000 – 5000
P(5000) = 20,000
Twenty thousand dollars is the maximum profit. Good luck. =)
References :
01.October, 2009 um 5:16 am
Assume you sold x quantities.
Profit = revenue – cost
P = (12x -0.001x^2) – (5000+2x)
P = 12x -0.001x^2 – 5000-2x
Maximize P
Differentiate with respect to x and set it equal to 0 and solve.
dP/dx = 12 – 2(0.001)x -2 =0
10-0.002x=0
-0.002x=10
x=10/0.002 =5,000
d^2P/dx^2 = -0.002 < 0, verifies the profit has been maximized.
To find the maximum profit, substitute x=5000 into R(x)-C(x)
P = 12x -0.001x^2 – 5000-2x
when x=5,000
P = 20,000 is the maximum profit.
References :