Total Differential to find slope of Curve

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July 29th 2009, 08:09 AM
Apache

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.
July 29th 2009, 08:29 AM
Danny

Quote:

Originally Posted by Apache

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 $\frac{dy}{dx}$ .
July 29th 2009, 10:27 AM
Apache

Thanks, so I would get something like this ?

$(-4x+4y+64)dx +((4x-8y+32)dy)$
July 29th 2009, 11:00 AM
Danny

Quote:

Originally Posted by Apache

Thanks, so I would get something like this ?

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

Yes, now isolate $\frac{dy}{dx}$ .
July 29th 2009, 11:23 AM
Apache

Think I’ve 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 ?