The cost and price functions for a new Internet search company are reasonably approximated by the following models:

$C(x) = 37 + 1.42x – 0.0067x^2 + 0.00011x^3$
$p(x) = 3.7 – 0.007x$

Where $x$ represents the number of searches (in millions) per month. Find the ideal number of searches this company can manage (to maximize profit) per month.

(A) Approximately 65 million searches per month

(B) Approximately 76 million searches per month

(C) Approximately 82 million searches per month

(D) Approximately 97 million searches per month

(E) None of the above

So I believe the function to be maximized would be the price function minus the cost function:

Profit function, $G(x) = p(x)-C(x) = 3.7 - 0.007x - 37-1.42x+0.0067x^2 -0.00011x^3$

Maximize function:

$G'(x) = 0 -0.007 -0-1.42 +0.0134 x -0.00033x^2 = 0$

I've been getting complex imaginary numbers here and I was wondering if I did something wrong or if the answer is actually none of the above. It seems like there should be real answers but I don't know how to explain why if there isn't.

  • $\begingroup$ The problem is that $p(x)< c(x)$ everywhere, so the equation you wrote down has no roots. Are you sure you copied the equations correctly? $\endgroup$ – rogerl Mar 7 '14 at 23:25
  • $\begingroup$ shouldn't you multiply the price by $x$ to get revenue first? $\endgroup$ – Maroon Mar 7 '14 at 23:34

As I mentioned in my comment, your equations seem to be right, but you don't have the correct function for optimization - with my method below, Wolfram Alpha gives an acceptable answer.

You want to maximize profit, and you have the price function and the cost function. To maximize profit, you optimize by subtracting cost from revenue, and optimizing the resulting function with appropriate restrictions (i.e. $x>0$). Notice that:

  • The price function gives the price for a single unit. Thus, to get revenue, we need to multiply the price function by $x$, so we need to use $x(3.7-0.007x)$ in the final equation, instead of $3.7-0.007x$.

  • Why can't we use the price function? Because the price function here is simply giving us the price of a single unit as a function of the number of units that are sold.

  • The cost function, however, is already implicitly (not in the mathematical sense) a function for total cost, since it takes into account the number of units we are selling. Thus, we don't need to modify it.

This means that we need to optimize $(3.7-0.007x)x-37-1.42x+0.0067x^2-0.000111x^3$, subject to appropriate constraints on $x$. You should get a reasonable result from this.

  • $\begingroup$ Ok so I did this and: Profit function, G(x) = (3.7−0.007x)x-37-1.42x+0.0067x^2-0.00011x^3 => 3.7x−0.007x^2-37-1.42x+0.0067x^2-0.00011x^3 => 2.28x-37+0.0003x^2-0.00011x^3 => Maximize function G'(x) = 2.28-0+0.0006x-0.00033x^2= 0 And then wolphram alpha gave me -82.217 and 84.035 Does this mean the answer is C. 82? $\endgroup$ – Brian Mar 8 '14 at 0:26
  • $\begingroup$ Which function were you maximizing? You need to use the maximize command on $G(x)$, not $G'(x)$. I get this: wolframalpha.com/input/… $\endgroup$ – Maroon Mar 8 '14 at 0:29
  • $\begingroup$ I was trying to find the maximum mathematically by deriving the function and setting it to zero. But I could be wrong, I'm teaching this to myself with the internet for this problem. $\endgroup$ – Brian Mar 8 '14 at 0:53
  • $\begingroup$ But Wolfram did give me the maximum and an answer so thanks for your help. :) $\endgroup$ – Brian Mar 8 '14 at 0:59
  • $\begingroup$ @Brian: you need to set $G'(x)$ equal to $0$, not $G(x)$ - I just used Wolfram Alpha since I was too lazy to work out the derivative and get the appropriate quadratic form to solve. Feel free to upvote/accept answers if they are useful to you. $\endgroup$ – Maroon Mar 8 '14 at 1:02

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