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Thread: 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,

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

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

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

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

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

    To check these are the maximum points,

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

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

    $\displaystyle UxxUyy-Uxy^{2}$

    = 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 = $\displaystyle \frac{dy}{dx}$

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

    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,

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

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

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

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

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

    To check these are the maximum points,

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

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

    $\displaystyle UxxUyy-Uxy^{2}$

    = 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 = $\displaystyle \frac{dy}{dx}$

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

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

    $\displaystyle (-4x+4y+64)dx +((4x-8y+32)dy)$
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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 ?

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

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

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

    $\displaystyle (MUx)(dx)+(MUy)(dy)$

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

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

    Rearranging gives

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

    Substitute equation for $\displaystyle MUx$ and $\displaystyle MUy$,

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

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

    Is that the expression for my slope ?
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