CALCULUS AND OPTIMIZATION TECHNIQUES

Economists spend considerable time trying to identify maximum and minimum points on a function.  For example, we assume consumers want to consume goods that maximize utility and that firms want to produce outputs that maximize profit and use inputs that minimize costs.  Calculus provides the mathematical tools for doing this.

1.         If y = f(x) is a continuous function (no weird kinks or gaps), then the derivative of y with respect to x (written as dy/dx) shows the change in y that will occur if there is an "infinitely small" change in x.  Geometrically, dy/dx is the slope of the function.

2.         For power functions such as y = axn the derivative dy/dx = naxn‑1

             As examples:

                   if  y = 7x3             using the above formula, think of a = 7 and n = 4
                                               dy/dx = 3(7)x3-1 = 21x
                   if y = -3x4             in this case a = -3 and n = 4
                                                dy/dx = 4(-3)x4-1 = -12x3
                   if y = 12x              a = 12 and n = 1 (the 1 is not written, but is understood)
                                                dy/dx = 1(12)x1-1 =  12x0 = 12  (any variable raised to an exponent of 0 = 1)
                   if y = 5                  think of this as y = 5x0
                                                dy/dx = 0(5)x0-1 = 0

 3.       The derivative of a sum is the sum of the derivatives.  Thus, if

             y = f(x) + g(x),              dy/dx = df(x)/dx + dg(x)/dx

             As examples:

                       if y = 2x - 4x2                                  dy/dx = 2 - 8x

                       if y = 50 - 3x + x2 - 6x3                   dy/dx = -3 + 2x - 18x2

4.     Higher ordered derivatives can be calculated by differentiating over and over again.
        As an example: if y = 3x2,

         the first derivative (dy/dx)                                  = 6x
         the second derivative  =  the derivative of 6x      = 6
         the third derivative  =  the derivative of 6            = 0

5.         Derivatives are slopes and slopes are zero at maximum or minimum points of a function.  Therefore, the "first order condition" for finding a maximum or minimum point of a function is finding where the first derivative (or slope) of the function is zero.  

            To determine whether the solution is a maximum or a minimum, "second order conditions" must be examined.  The second order condition for a maximum is that the second derivative must be negative while the second order condition for a minimum is that the second derivative must be positive.

 

SAMPLE PROBLEMS

 1.         Suppose y = 800x - 5x2          To find the maximum or minimum point: 

            a.  differentiate.... dy/dx = 800 - 10x
            b.  find where derivative equals zero...
                      800 - 10x = 0
                               10x = 800
                                   x = 80
            c.  check second-order condition...          d2y/dx2 = -10
                 Because the second derivative is negative, the point is a maximum.

 2.         Suppose y = 500 - 1200x + 135x2 - 2x3   To find maximum or minimum ponits:

             a. differentiate...dy/dx = -1200 + 270x - 6x2
             b. find where the derivative equals zero...
                   -1200 + 270x - 6x2 = 0
                           200 - 45x + x2 = 0
                            (40 - x)(5 - x) = 0
                             x = 40   or  x = 5

             c.  check second-order condition....d2y/dx2 = 270 - 12x

                           If x = 40, the second derivative = 270 - 12(40).
                           Because this is negative, we must have a maximum at x = 40.
                           If x = 5, the second derivative = 270 - 12(5).
                           Because this is positive, we must have a minimum at x = 5.