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Math Help - Optimisation in Two dimensions

  1. #1
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    Optimisation in Two dimensions

    I have to find all the critical points and their nature from the following scenario:

    There are 2 types of bread- plain and gluten free. Weekly production is represented by x hundred plain loaves and y hundred gluten free. Plain sells for 90 cents each and gluten free for $5.20 each. Weekly production costs are given by:
    10x^3 + 20y^3 -10xy

    I have started by putting x and y as functions:
    fx= 30x^2 - 10y
    fy= 60y^2 - 10x

    Making them equal zero gave me
    x= 6y^2
    y= 3x^2

    How would you solve x=6(3x^2)^2 to find the critical points?
    What would I have to do after?

    Thanks
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  2. #2
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    Quote Originally Posted by Pandora View Post
    I have to find all the critical points and their nature from the following scenario:

    There are 2 types of bread- plain and gluten free. Weekly production is represented by x hundred plain loaves and y hundred gluten free. Plain sells for 90 cents each and gluten free for $5.20 each. Weekly production costs are given by:
    10x^3 + 20y^3 -10xy

    I have started by putting x and y as functions:
    fx= 30x^2 - 10y
    fy= 60y^2 - 10x

    Making them equal zero gave me
    x= 6y^2
    y= 3x^2

    How would you solve x=6(3x^2)^2 to find the critical points?
    What would I have to do after?

    Thanks
    x = 6(3x^2)^2

    x = 6(9x^4)

    x = 54x^4

    0 = 54x^4 - x

    0 = x(54x^3 - 1)

    0 = x[(\sqrt[3]{54}x)^3 - 1^3]

    0 = x[(3\sqrt[3]{2}x)^3 - 1^3]

    0 = x(3\sqrt[3]{2}x - 1)(9\sqrt[3]{4}x^2 + 3\sqrt[3]{2}x + 1 )


    So x = 0 or 3\sqrt[3]{2}x - 1 = 0 or 9\sqrt[3]{4}x^2 + 3\sqrt[3]{2}x + 1 = 0.

    Solve for x in each case.
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  3. #3
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    That was very helpful, thank you! Would you mind telling me how to work out the nature of those 3 critical points also? I know you have to use the Hessian matrix, however, I am pretty terrible at partial derivatives.
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