Question

The cost of running a small factory is modeled by the function where x is the number of routers (in hundreds) sold weekly and C is in dollars. C(x)= x^3-3x^2+18x

The company sells routers for 42 dollars each. Write the equation for revenue as a function of number sold. (Hint: revenue = number sold * price.) R(x) =

Answer 1

I fail to see the complexity of this problem.
'revenue = number sold * price'
revenue R(x)
number sold 'x is the number of routers (in hundreds) sold weekly'
price 'for 42 dollars each'
therefore it should be simply 42*x?

Answer 2

You are so right, I forgot to attach the bottom half of the problem, which i had trouble with. The rest of the question is 1. Find the number of routers sold which will maximize weekly profit. (Hint: profit = revenue minus cost). 2. What is the company's maximum profit? I appreciate the help!

Answer 3

Let profit be a function of numbers sold: P(x)
P(x) = R(x) - C(x)
P(x) = 42x - x^3 + 3x^2 - 18x
P(x) = -x^3 + 3x^2 + 24x
Now find when P'(x) = 0; Is when P(x) at extrema
0 = -3x^2 + 6x + 24
...
x = 4 or -2
now find 2nd order derivatives at these 2 values to determine nature of extrema
P''(x) = -6x + 6
P''(4) = -18 < 0
-2 is an unrealistic solution to the problem, therefore there is no need to examine it further

Now at weekly production of 4 the weekly profit is maximized

profit at that point: P(4) = -4^3 + 3*4^2 + 24*4 = -64 + 48 + 96 = 80

Answer 4

thank you so much!