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Course: MOREY 4808, Fall 2009
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higher-order 1 Consider derivatives April 12, 2007 higher.tex We have de...ned the cost function, c = c(x), and the marginal cost function as mc(x) = c0 (x). Since mc(x) is itself a function, one could take its derivative. E.g. dmc(x) dx mc0 (x) = c00 (x) d dc(x) dx dx d2 c(x) dx2 I just introduced some new notation. While mc0 (x) is the ...rst-order derivative of the marginal cost function with respect to x,...

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higher-order 1 Consider derivatives April 12, 2007 higher.tex We have de...ned the cost function, c = c(x), and the marginal cost function as mc(x) = c0 (x). Since mc(x) is itself a function, one could take its derivative. E.g. dmc(x) dx mc0 (x) = c00 (x) d dc(x) dx dx d2 c(x) dx2 I just introduced some new notation. While mc0 (x) is the ...rst-order derivative of the marginal cost function with respect to x, it is the second-order derivative of the cost function with respect to x. The following are all equivalent ways or representing a second-order derivative d dc(x) d2 c(x) dx c00 (x) dx dx2 Here the superscript 2 denotes that it is a second derivative, not that we are "squaring" something. If one can take second-derivatives, one can also take third and higher-order derivatives. E.g. d4 c(x) c0000 (x) dx4 is the fourth-order derivative of the cost function with respect to x. One can take a derivative of a function, no matter what its origins. 1 Consider the following cost function: c(x) = x3 12x2 + 60x c(x) 400 300 200 100 0 0 2.5 5 7.5 x 10 2 Describe, in words, the properties of this cost function. Find this ...rm' marginal cost function: s mc(x) = c0 (x) = 3x2 24x + 60 Note that marginal cost is also a function of x. What does the marginal cost function look like? One can determine this by graphing the function and/or looking at the second-order and higher derivatives of the cost function (the ...rst-order and higher derivatives of the marginal cost function). mc(x) = c0 (x) = 3x2 mc(x) 100 75 50 25 0 2.5 5 7.5 x 10 24x + 60 > 0 8x 0? Note that mc(0) = 60. 3 Is marginal cost always positive? How might one tell? One could visually examine the marginal cost function and see if it is always positive. It is a bit di cult to tell on the previous Figure, so let' look at the s marginal cost function in the range 3:75 to 4:25 (the range where it takes is minimum value) mc(x) 12.18 12.15 12.13 12.1 12.07 12.05 12.02 12 3.75 3.875 4 4.125 x 4.25 It takes its minimum value, 12, at x = 4. We can con...rm this by plugging 4 into the marginal cost function to get mc(4) = c0 (4) = 3(4)2 24(4) + 60 = 12. So yes, marginal cost is always positive. What would it mean for a "cost" function if at some level of output, marginal cost was negative? Consider more the function, mc(x). We can take its derivative wrt x mc0 (x) = c00 (x) = 6x mc'(x) 25 24 12.5 0 0 -12.5 2.5 5 7.5 x 10 4 ) mc0 (x) < 0 if x < 4 and mc0 (x) > 0 if x > 4 What does the above tell us? Marginal cost is 60 when x = 0 then declines (mc0 (x) < 0) until x = 4 and then rises forever. Its minimum value is at x = 4 and mc (4) = 3(4)2 24 (4) + 60 = 12, so marginal cost is never negative. 5 Continuing to assume that c(x) = x3 12x2 + 60x y 400 300 200 100 0 0 2.5 5 7.5 x 10 Using derivatives,determine the of shape the average cost function. If c(x) = x3 ac(x) = \$ 100 75 50 25 0 2.5 5 7.5 x 10 12x2 + 60x 12x + 60 c(x) = x2 x The shallow u-shape is ac(x), the deeper-shaped u is mc(x). Note that ac(0) = 60. In addition 6 ac0 (x) = 2x 12 ) ac0 (x) < 0 if x < 6 and ac0 (x) > 0 if x > 6 What does the above tell us? Average cost is 60 when x = 0 then declines until x = 6 and then rises forever. Its minimum value is at x = 6. At what level of x if any does ac(x) = mc(x)? Solve ac(x) = x2 12x + 60 = 3x2 24x + 60 = mc(x) for x There are two solutions, one of which we already identi...ed: x = 0 and x = 6. Average cost equals marginal cost when average cost is minimized, also when x = 0. If 0 < x < 6 marginal cost is less than average cost, so the marginal cost is pulling the average cost down. If x > 6 marginal cost is greater than average cost, so the marginal cost is pulling the average cost up. 7 Let' see if we can derive the general result, one that applies to all cost s functions, not just the above example. Since c(x) x One can apply the quotient rule to determine ac(x) = ac0 (x) = d( c(x) ) c0 (x)x c(x)1 1 0 x = = c (x) x x2 x c(x) 1 = [c0 (x) x x ac(x)] The stu in brackets is marginal cost minus average cost, so is positive if marginal cost is greater than average cost, and negative if average cost is greater than marginal cost. That is ac0 (x) = dac(x) 7 0 as [c0 (x) x ac(x)] 7 0 8 In words, ........ In general, what does some function f (x) look like if f 0 (x) > 0 and f 00 (x) > 0? the function increases at an increaseing rate (slope becomes more positive as x incr f 0 (x) > 0 and f 00 (x) < 0? the function increases at a decreasing rate (slope becomes less positive as x increase f 0 (x) < 0 and f 00 (x) > 0? the function decreases at a decreasing rate (slope becomes less negative as x increas f 0 (x) < 0 and f 00 (x) < 0? the function decreases at an increasing rate (slope becomes more negative as x incr What would a graph of the functions with the following properties look like? f 0 (x) < 0 and f 00 (x) > 0 and f 0 (x) < 0 and f 00 (x) < 0 look like. Have them break into groups and ...gure it out. 9 Start by considering the following function. Consider the function f (x) = 1 :1ex = :1e y 10 x 7.5 5 2.5 0 1.25 2.5 3.75 x 5 Take the derivatives d x ) = 0:1e 1:0x dx (:1e x Dx (:1e ) = 0:1e 1:0x < 0 Dxx (:1e x ) = 0:1e 1:0x > 0 Dxxx (:1e x ) = 0:1e 1:0x < 0 Note that one could do this forever, and all that would happen is that the signs switch from negative to positive to negative ..... So, f (x) = 1 :1ex = :1e x has the property that f 0 (x) < 0 and f 00 (x) > 0 :1ex 2.5 x 5 now consider another function , f (x) = -5 -2.5 0 -2.5 -5 -7.5 -10 -12.5 y 0 Dx ( :1ex ) = 0:1ex < 0, Dxx ( :1ex ) = 0:1ex < 0 10
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