The ability to use calculus to find minima and maxima is very useful in
many areas of study. Economics is no exception. If we can maximize our profit
and minimize our costs, our business goals can approach the optimum. Below
is a chart of economic terms and formulas that will allow us to solve some
economics problems and make use of our derivative skills:
Total Cost  C(x) 
Marginal Cost 
C'(x) 
Average Cost  
Price Function  p(x) 
Revenue Function  R(x) = x p(x) 
Marginal Revenue  R'(x) 
Profit Function  P(x) = R(x)  C(x) 
Marginal Profit  P'(x) = R'(x)  C'(x) 
We see that whenever we find Marginal Cost or Marginal Revenue, or Marginal Profit, we are finding the instantaneous rate of change or derivatrive. As we often want to minimize average cost in business, it becomes important to business problem solving to recognize that:
Example Problem:
Given the cost function: 
Given Problem, #8, Lesson 4.7 
To find average cost, we know that we need to use the formula: .  
To find the marginal cost, we use the formula: C'(x)  
We now move on to part (b) 
We plot the graph of the average cost (red), and the graph of the marginal cost (green). By observation, it appears that the value of x (number of items produced) where the two graphs intersect is about x = 140. 
We now work on part (c)

Because of our principle that: "If the average cost is a minimum, then marginal cost = average cost," we set the average cost formula equal to the marginal cost formula and attempt to solve. 
> solve(0=.0008*x^3.09*x^2339,x);
11.52974818  54.70830486 I, 11.52974818 + 54.70830486 I, 135.5594964 
Here we need a computer algebra system to find the solution to this cubic equation. We see there are two imaginary solutions and one real solution. The answer is approximately, This agrees with our visual estimate above. 
