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Math Help - Pls check my work – bivariate optimization

  1. #1
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    Pls check my work – bivariate optimization

    I’ve just finished my textbook, “Forgotten Calculus”. The last chapter is on multivariable calculus. Disappointingly, while there were some “pure math” problems on finding and classifying relative extrema, there were no real-world applications of this technique, i.e optimization problems.

    So I decided to make one up, or rather base it on a marginal profit problem in the text. So what I’ve done below is not an “assigned” question per se. So the book has no answer for me to check myself by.

    I would therefore ask any of you kindly, mentoring types out there to tell me if I am going about this right or wrong (I suspect the latter) and if need be correct me. To make it easier I show pretty much all of my work, although it does make this post pretty long.

    I also need some guidance on interpretation.

    Lastly, if anyone can recommend any sources which have well-explained, worked examples in multivariate optimization, I’d be grateful. I don't think I need (or have time for) a whole book on the subject; 5-10 examples would help a lot.

    The premise

    Total annual profit z from a small “summer hobby” farm is determined as

    z = f(x,y) = 10x^3 + 20y^2 – 10xy

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

    This much came from the text.

    The question and strategy summary

    I’d like to find out which combination(s), if any, of ad spending and acres maximizes (or minimizes) profit.

    I will attempt to do so by trying to locate and evaluate the critical points, and hope they make some sense.

    By the way: as a reference point, I evaluate the original function at values given, i.e

    If

    f(x,y) = 10x^3 + 20y^2 – 10xy

    then

    f(10, 5) = 10 (10)^3 + 20(5)^2 – 10 (10) (5)
    = 10,000 + 500 - 500
    =10,000

    Find critical points

    Start by taking the partials f[subscript]x and f[subscript]y

    f[subscript]x = 30x^2 – 10y
    f[subscript]y = 40y – 10x

    Next, solve for x and y by setting each partial equal to 0, and solving as a “system of equations”, if this is the right term.

    0 = 30x^2 – 10y
    0 = 40y – 10x

    Rearrange the bottom equation…
    0= 30x^2 - 10y
    0= -10x + 40 y

    Put them into a common format…
    0 = 30x^2 – 0x – 10y
    0 = 0x^2 – 10x + 40y

    Multiply the top equation by 4 to get the “y”s to cancel out when I then “add” the two equations together….

    0 = 120x^2 – 0x – 40y
    0 = 0x^2 – 10x + 40y
    ___________________
    0 = 120x^2 – 10x +0

    Use the quadratic formula (done on a calculator, work not shown) to solve for x

    Point 1: x = .08333 (displayed in this explanation showing only 3 threes, but in calculations, using 6 threes)

    Point 2: x = 0

    Solve for y at Point 1 by plugging the Point 1 x value into one of the first partials set equal to 0

    f[subscript]x = 30x^2 – 10y

    0 = 30x^2 – 10y
    0 = 30(.08333)^2 – 10y
    0 = .208333 – 10 y
    .208333 = 10y
    Point 1: y = 48
    So Point 1 = (.08333; 48; z)


    Solve for y at point 2 using the same method, only using the Point 2 x value

    f[subscript]x = 30x^2 – 10y
    0 = 30x^2 – 10y
    0 = 30 (0) – 10y
    0 = 10y
    Point 2: y = 0
    So Point 2 = (0; 0; z)


    Find z for Point 1

    Plug the Point 1 x and y values into the original function

    z = 10x^3 + 20y^2 – 10xy
    = 10 (.08333)^3 + 20(48)^2 - 10 (.08333)(48)
    = .005787 + 46,080 – 39.999

    Point 1: z = 46,040
    So Point 1 = (.08333; 48; 46,040)


    Find z for Point 2 using the same method, only plugging in Point 2 x values

    z = 10x^3 + 20y^2 – 10xy
    = 0 + 0 -0
    Point 2: z = 0
    So Point 2 = (0; 0; 0)


    An aside

    At this point, before properly classifying the critical points, I was thinking ahead: “Ok, this makes sense – Point 1 is going to be a maximum, at which the maximum profit is 46,060, which is within the same order of magnitude as the profit under the premise conditions, and therefore seems plausible; while Point 2, showing 0 profit when there is no ad spending and no acres to far, makes perfect sense as a minimum”

    As you’ll see below, I was wrong…

    Classifying the critical points

    Let D (x,y) = [ ( f[subscript]xx ) ( f[subscript]yy ) ] - (f[subscript]xy )^2

    From above, for reference:

    f[subscript]x = 30x^2 – 10y
    f[subscript]y = 40y – 10x

    So

    f[subscript]xx = 60x
    f[subscript]yy = 40
    f[subscript]xy = 10

    So

    D= [ (60x) (40) ] - (10)^2
    D= 240x – 100

    For Point 1:

    D = 240(.08333) - 100
    D = 20 – 100
    D = - 80
    This is negative; thus Point 1 is a saddle point


    For Point 2:
    D = 240(0) - 100
    D = – 100
    This is negative; thus Point 2 is also a saddle point


    Concluding questions

    First: Have I conceived and / or executed the solution wrong?

    Second: I must say this is somewhat unsatisfying. Assuming I’ve done this right, I’m not sure what to make of the answer. I know what a drawing of a saddle looks like but frankly don’t understand what it means in this case – a profitability problem with no minima or maxima, only two saddle points. How would you plan an optimal business strategy based on this?

    If you’ve read this far, thanks for your attention.

    Regards,

    lingyai
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by lingyai
    I’ve just finished my textbook, “Forgotten Calculus”. The last chapter is on multivariable calculus. Disappointingly, while there were some “pure math” problems on finding and classifying relative extrema, there were no real-world applications of this technique, i.e optimization problems.

    So I decided to make one up, or rather base it on a marginal profit problem in the text. So what I’ve done below is not an “assigned” question per se. So the book has no answer for me to check myself by.

    I would therefore ask any of you kindly, mentoring types out there to tell me if I am going about this right or wrong (I suspect the latter) and if need be correct me. To make it easier I show pretty much all of my work, although it does make this post pretty long.

    I also need some guidance on interpretation.

    Lastly, if anyone can recommend any sources which have well-explained, worked examples in multivariate optimization, I’d be grateful. I don't think I need (or have time for) a whole book on the subject; 5-10 examples would help a lot.

    The premise

    Total annual profit z from a small “summer hobby” farm is determined as

    z = f(x,y) = 10x^3 + 20y^2 – 10xy

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

    This much came from the text.

    The question and strategy summary

    I’d like to find out which combination(s), if any, of ad spending and acres maximizes (or minimizes) profit.
    If z=1000000, and y=5, what is the profit and how does that compare with
    the profit at any critical points you may have found.

    Now what about z=1000000000, y=5?

    RonL
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  3. #3
    Junior Member
    Joined
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    Quote Originally Posted by CaptainBlack
    If z=1000000, and y=5, what is the profit and how does that compare with
    the profit at any critical points you may have found.

    Now what about z=1000000000, y=5?

    RonL
    Sorry Captain, I think I've missed your point.

    Profit being Z, if Z = 1,000,000, then profit = 1,000,000. Likewise If Z = 1 bn.

    I'm not trying to be deliberately thick here. I'm trying, untutored, to expand beyond the limits of a text which I suspect is being too light in what seems to be an important area. Having only been doing multivariate calculus for 2 days, I feel that in doing so the water's just below my nostrils. Could you please spell out your message more directly?

    Is my strategy wrong? Is my math wrong? Is the question wrong?

    lingyai
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  4. #4
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    From my understanding of your problem, you have found the internal maxima and minima, which turn out to be saddle points, tbh I would expect this. You haven't defined a limit, so how could you expect to calculate an extremum (global maximum) when the higher the variables are, the higher the result is? It would be infinite, thus no mathematical solution for a maximum.

    You have to define some limits, saying that there is a maximum acres, and a maximum spendature otherwise it'll be infinite and you're maths will never calculate a maximum. In your example these limits might be implied by finance and available farming terrain.

    Mathematically, consider y=x^2

    If you were asked whats the absolute maximum value anywhere along that line, would you be able to answer? No, not unless your answer was infinite.

    If you were asked whats the absolute maximum value between the points 0<x<10 would you be able to answer? Yes, and you would expect the maximum value to be at the boundary, in this case x=10, y=100.

    The same situation applies to multivariable calculus. If theres no boundary, most cases you'll never find a global maximum/minimum.
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  5. #5
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    ......... wow that was kinda dumb of me to reply to a post this old. But it was in my search and maybe it will help someone.

    Sorry.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Makka View Post
    ......... wow that was kinda dumb of me to reply to a post this old. But it was in my search and maybe it will help someone.

    Sorry.
    Of course I should have replied a long time ago, explaining what my questions were an attempt to point out to the OP.

    RonL
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