Kline Calculus Problems - Simple Derivatives and Marginal Cost

  1. Apr 5, 2012 #1
    I had questions on 2 Problems in the Text:

    1. The total cost C of producing x units of some item is a function of x. Economists use the term marginal cost for the rate of change of C with respect to x. Suppose that:

    C = 5x^2 + 15x + 200

    What is the marginal cost when x = 15? Would this marginal cost be the cost of the 16th unit?

    (((I understand how to take the derivative and find dC/dx. However, I am unsure as to why this represents to cost of the 16th unit. Also, can someone explain to me in simple terms what the derivative of C represents?)))

    2. Using the definition of marginal cost in the preceding exercise, suppose that the cost C of producing x units of a toy is C = 3x^2 - 4x + 5. What is the marginal cost at any value of x? Would the marginal cost necessarily increase with x in any realistic situation?

    (((Why doesn't the marginal cost always increase with x in any realistic situation?)))

  2. jcsd
  3. Apr 5, 2012 #2
    Hey ghostskwid and welcome to the forums.

    The derivative means the instantaneous rate of change of something with respect to another. In this case it represents how C changes with x at that point. Think of looking at the slope of the function C between a point x and x + dx and what happens is that dx gets smaller and smaller to goes to zero but isn't zero! It's a weird thing to understand but that's the best way to describe it.

    Basically in this context if the slope is increasing then the cost is increasing for every x which means there will be a relative increase in cost to produce more stuff and if it decreases then it will cost relatively less.

    The thing that businesses want to do is create more things at the cheapest possible rate which means that if we have any dC/dx where it is negative then this means that the businesses can create or produce more things without having to spend as much for each new piece of stuff (in other words it's less per unit to produce more stuff than it is to produce the existing stuff).

    Think about a factory creates say a lot of cars or something on an assembly line. They have to pay for running the assembly line, wages, and all that stuff as well as for the materials but once they produce enough to cover things like wages and operating the factory, then it won't cost them as much to produce anything more and this is what businesses with factories want because they will sell their stuff at the same price usually which means they make a lot more profit when they make more stuff if the dC/dx is negative.

    By finding the turning point where dC/dx is zero at a minimum, this says for the business what's the best amount to produce to maximize profit in one sense.

    This should help you think about the second question.
  4. Apr 5, 2012 #3
    Thanks that helps.

    Why is the marginal cost at 15 the actual cost of unit 16?

    Also in the second problem could the marginal cost ever be negative?
  5. Apr 5, 2012 #4

    It seems to be that (1) asks the following: the marginal cost at x = 15 is [itex]C'(15)=10\cdot 15+15=165[/itex] .

    On the other hand, the cost of the 16th product seems to be C(16)-C(15) = the cost of making 16 items minus the cost of making 15 items, and this gives 170, so no.


    Disclaimer: the person that wrote the above answer is a pure mathematician and thus his messing with mathematicial economics and/or financial stuff must be taken with due care.
  6. Apr 5, 2012 #5
    I'm not a pure mathematician: my background is in computer programming and I will graduate this year with a double major in statistics and applied mathematics.
  7. Apr 5, 2012 #6


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    chiro, I believe Don Antonio was referring to himself.

    ghostskwid, yes, the marginal cost of the 16th item is just the cost of that item. Since your function, C(x), gives the total cost of manufacturing x items, the marginal cost of the 16th item is C(16)- C(15).

    Since you refer to both "simple derivatives" and "marginal cost" in the title of this thread it might be good to point out that the marginal cost of the "x" item is
    [tex]C(x+ 1)- C(x)= \lim_{h\to 1}\frac{C(x+h)- C(x)}{h}[/tex]
    while the derivative is
    [tex]\lim_{h\to 0}\frac{C(x+h)- C(x)}{h}[/tex]

    Of course, the "h" going to 1 in the denominator raises much less theoretical issues than it going to 0!
  8. Apr 5, 2012 #7
    Thankyou HallsOfIvy for that. Hopefully the OP will reply so that everything gets cleared up.
  9. Apr 5, 2012 #8
    I get 165 when calculating the marginal cost at $15 and 170 when calculating the cost of the 16th product ((C(16) - C (15))). Kline reports the answer as yes it will be.

    Could you explain your tex block. How do they differ? I thought the marginal cost was simply the derivative of the function. Thanks
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