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Math Help - Significance of a second partial in marginal analysis?

  1. #1
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    Significance of a second partial in marginal analysis?

    I'm doing a bivariate calculus problem from a textbook. I can reach all the answers, but am unsure about something -- what does an evaluated second partial derivative mean in a marginal analysis context?

    Here's the question. (I won't show all the work in reaching the answers, as they are not problematic or controversial.)

    Total annual profit f(x,y) is determined as

    f(x,y) = 10x^3 + 20y^2 - 10xy,

    when x = ad spending = 10 and
    y = acres farmed = 5.

    I do understand:
    that f(10,5) = 10,000 = total annual profit;
    that f[subscript]x = 30x^2 - 10y; and
    that f[subscript]x (10,5) = 2950 = the approximate additional annual profit to be gained from increasing ad spending by one unit (i.e. from 10 to 11).

    What I DON'T understand is something the author has us calculate, but which we never end up using in the analysis, and which she never comments on. Specifically, she asks for

    f[subscript]xx (i.e the second partial) = 60x and
    f[subscript]xx (10,5) = 600.

    I can reach these answers, but I don't understand, what does 600 actually mean here?

    I'd be grateful for any advice.
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  2. #2
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    The second derivative and second partial derivatives are important in optimization problems because they tell you whether a critical point (where the first derivatives equal zero) is a local maximum, minimum or indeterminate. For example, for the function x^3



    the critical point at x = 0 is neither a local min or max. The second derivative is 6x = 0. If the second derivative were negative, the critical point would be a local max and if it were positive, a local min. The reason is that the second derivative tells you how the first derivative (the slope) is changing. If the second derivative is negative, the slope is decreasing, so you have a hill with a maximum. If the second derivative is positive, the slope in increasing, so you have a valley with a minimum. For the function x^3, the slope is decreasing and then increasing around the critical point, so you have neither a hill nor a valley.

    It two dimensions, the second partials have the same function. For example here



    is another case where there is a critical point that is neither a min or a max. But in two dimensions, you have to look at the 2x2 matrix of second partial derivatives called the Hessian matrix.
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  3. #3
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    Jake,

    Thank you for your answer.

    Permit two follow-on questions:

    1) I am familiar with the 2nd derivative test as one way to find extrema in, for example, optimization problems. (It is offered in my text as one of three methods for doing so; I personally prefer to use the first derivative test). But I'm struggling to see what, specifically in the problem I related (which is not an optimzation problem, at least not on the surface), what the value 600 means. (Forgive me if I'm a little slow on the uptake; I'm teaching myself calculus and am only average in the math leagues).

    2) How did you generate those 3-d graphs? They would come in very handy as I do multivariate problems, but seem beyond my TI 83 Plus.

    Thanks again, and regards,

    Lingyai
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  4. #4
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    Lingyai,

    To be really familiar with the second derivative test, you should be familiar with what the second derivative means. I went through that test because it shows how the second derivative is used and what it means. Let's go back to the definition of the second partial.

    \displaystyle{ \lim_{\Delta x \to 0} \frac{ f_x (x+\Delta x,y) - f_x (x,y) }{\Delta x} = f_{xx} (x,y)}

    means for small \Delta x that

    f_x (x+\Delta x,y) - f_x (x,y) \approx f_{xx} (x,y) \Delta x .

    So if f_{xx}(x,y) = 600 , the change in the first partial is

    f_x (x+\Delta x,y) - f_x (x,y) \approx 600 \Delta x .

    That is, the second partial just tells you how a slope (the first partial) changes with small changes in x. Taken literally, that's all there is. But the applications are many. The second derivative test is a very important application for economics, which your problem comes from.

    I didn't generate the graphs. They are links to graphics on a web page.

    Enjoy.

    Jake
    Last edited by JakeD; May 30th 2006 at 04:03 AM.
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