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Math Help - Total Differential to find slope of Curve

  1. #1
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    Total Differential to find slope of Curve

    An individuals utility depends upon the consumption of goods x and y,

    U=2x ^{2}+4xy-4y^{2}+64x+32y-14

    I have to find the utility maximisation of x and y,

    \frac{\partial y}{\partial x}=-4x+4y+64 (1)
    \frac{\partial x}{\partial y}=4x-8y+32 (2)

    Multiply (1) by 2 = 128+8y-8x (3)
    32-8y+4x (4)

    (3)-(4), after rearranging and substitution,
    x = 8, y = 8

    To check these are the maximum points,

    \frac{\partial ^2y}{\partial x^2}=-4
    \frac{\partial ^2x}{\partial y^2}=-8

    Find cross partial derivitives,
    Uxx = -4
    Uyy = - 8
    Uxy = 4

    UxxUyy-Uxy^{2}

    = 16 > 0
    3 conditions meet = maximum

    I then have to use the total differential of the function to find an expression for the slope of any curve.

    Slope of curve = \frac{dy}{dx}

    du=\frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy=0

    This is where i become a little stuck as to what I should do next.
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  2. #2
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    Quote Originally Posted by Apache View Post
    An individuals utility depends upon the consumption of goods x and y,

    U=2x ^{2}+4xy-4y^{2}+64x+32y-14

    I have to find the utility maximisation of x and y,

    \frac{\partial u}{\partial x}=-4x+4y+64 (1)
    \frac{\partial u}{\partial y}=4x-8y+32 (2)

    Multiply (1) by 2 = 128+8y-8x (3)
    32-8y+4x (4)

    (3)-(4), after rearranging and substitution,
    x = 8, y = 8

    To check these are the maximum points,

    \frac{\partial ^2y}{\partial x^2}=-4
    \frac{\partial ^2x}{\partial y^2}=-8

    Find cross partial derivitives,
    Uxx = -4
    Uyy = - 8
    Uxy = 4

    UxxUyy-Uxy^{2}

    = 16 > 0
    3 conditions meet = maximum

    I then have to use the total differential of the function to find an expression for the slope of any curve.

    Slope of curve = \frac{dy}{dx}

    du=\frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy=0 (**)

    This is where i become a little stuck as to what I should do next.
    Put in \frac{\partial u}{\partial x} and \frac{\partial u}{\partial y} from (1) and (2) (I think this is what you meant) into (**) and solve for \frac{dy}{dx} .
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  3. #3
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    Thanks, so I would get something like this ?

    (-4x+4y+64)dx +((4x-8y+32)dy)
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  4. #4
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    Quote Originally Posted by Apache View Post
    Thanks, so I would get something like this ?

    (-4x+4y+64)dx +((4x-8y+32)dy)
    Yes, now isolate \frac{dy}{dx} .
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  5. #5
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    Think Iíve cracked it,

    \frac{\partial y}{\partial x}=MUx

    \frac{\partial x}{\partial y}=MUy

    (MUx)(dx)+(MUy)(dy)

    \frac{dU}{dx}=MUx\frac{dx}{dx}+MUy\frac{dy}{dy}

    = MUx + MUy \frac{dy}{dx} = 0

    Rearranging gives

    -\frac{dy}{dx}=\frac{MUx}{MUy}

    Substitute equation for MUx and MUy,

     = -\frac{dy}{dx}=\frac{-4x+4y+64}{4x-8y+32}

     = -\frac{dy}{dx}= 2 -x -0.5y

    Is that the expression for my slope ?
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