Suppose you have a demand function p  80  0.01x and a cost function C ( x)  25 x  5000 , where x is the number of units demanded. a. Find the profit function. b. Find and interpret the marginal profit at 2000 units. To solve this problem, you’ll need to find the revenue function and then use it to get the profit function. Find the Revenue Function R(x) The revenue function is related to the units x and price p by R  xp . Since we already have the demand function solved for p, let’s substitute it to get R(x): R( x)  x  80  0.01x 
 80 x  0.01x 2
The function is a function of x since p no longer appears in the formula. Find the Profit Function P(x) The profit function P(x) can be found by realizing that profit = revenue – cost or P( x)  R ( x)  C ( x) If we put in the expressions for revenue and cost, we get P( x)   80 x  0.01x 2    25 x  5000 
 0.01x 2  55 x  5000
The graph of this function is shown below. 
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x units
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Find the Marginal Profit at x = 2000 To find the marginal profit, we need to take the derivative of the profit function. This is a fairly easy derivative: P( x)  0.02 x  55 The marginal profit at x  2000 is P(2000)  0.02  2000   55  15 To interpret this number, we need to realize that the marginal revenue is simply the instantaneous rate of change of the profit. Or we can think of it as the slope of a tangent line to the profit function at x  2000 . In either case, a rate has units of dollars per unit. The marginal revenue is thus 15 dollars per units. This means that an increase in production from 2000 units to 2001 units will lead to an increase in profit of $15. Since increasing production leads to an increase in profit, the increase would be a good idea. Note that the profit function is increasing at x  2000 which is consistent with the marginal profit being positive.