# Thread: Total Differential to find slope of Curve

1. ## Total Differential to find slope of Curve

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

$\displaystyle U=2x$$\displaystyle ^{2}+4xy-4y^{2}+64x+32y-14 I have to find the utility maximisation of x and y, \displaystyle \frac{\partial y}{\partial x}=-4x+4y+64 (1) \displaystyle \frac{\partial x}{\partial y}=4x-8y+32 (2) Multiply (1) by 2 = \displaystyle 128+8y-8x (3) \displaystyle 32-8y+4x (4) (3)-(4), after rearranging and substitution, x = 8, y = 8 To check these are the maximum points, \displaystyle \frac{\partial ^2y}{\partial x^2}=-4 \displaystyle \frac{\partial ^2x}{\partial y^2}=-8 Find cross partial derivitives, \displaystyle Uxx = -4 \displaystyle Uyy = - 8 \displaystyle Uxy = 4 \displaystyle UxxUyy-Uxy^{2} = 16 > 0 3 conditions meet = maximum I then have to use the total differential of the function to find an expression for the slope of any curve. Slope of curve = \displaystyle \frac{dy}{dx} \displaystyle du=\frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy=0 This is where i become a little stuck as to what I should do next. 2. Originally Posted by Apache An individuals utility depends upon the consumption of goods x and y, \displaystyle U=2x$$\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 $\displaystyle \frac{dy}{dx}$ .

3. Thanks, so I would get something like this ?

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

4. Originally Posted by Apache
Thanks, so I would get something like this ?

$\displaystyle (-4x+4y+64)dx +((4x-8y+32)dy)$
Yes, now isolate $\displaystyle \frac{dy}{dx}$ .

5. Think I’ve 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 ?