# Partial Differentiation stationary points

• Jan 15th 2011, 09:53 AM
ramdrop
Partial Differentiation stationary points
hey im pretty stuck with this, I've got quite far though, I think..

f(x,y) = $1/3x^3 + 1/3y^3 - 4xy^2 +15y$

then df/dx = $x^2 - 4y^2$
df/dy = $y^2 - 8xy + 15$

Then to find the stationary points, we must equate the differentials to 0, and I get to the 4 points:

(2y, -15), (2y, 8x+15), (-2y, -15), (-2y,8x+15)

Then by using the Hessian matrix

H = fxx fxy
fyx fyy

Where fxx = 2x, fyy = 2y - 8x, fyx = -8, fxy = 8y

Then the determinant is $4xy - 16x^2 + 64y$

Then you put in the x and y co-ordinates and determine whether is a maxima or minima as < or > 0.

But I don't have any numbers...so if anybody can help or see where I've gone wrong

Thanks
• Jan 15th 2011, 10:13 AM
Pranas
Quote:

Originally Posted by ramdrop
hey im pretty stuck with this, I've got quite far though, I think..

f(x,y) = $1/3x^3 + 1/3y^3 - 4xy^2 +15y$

then df/dx = $x^2 - 4y^2$
df/dy = $y^2 - 8xy + 15$

Then to find the stationary points, we must equate the differentials to 0, and I get to the 4 points:

(2y, -15), (2y, 8x+15), (-2y, -15), (-2y,8x+15)

Then by using the Hessian matrix

H = fxx fxy
fyx fyy

Where fxx = 2x, fyy = 2y - 8x, fyx = -8, fxy = 8y

Then the determinant is $4xy - 16x^2 + 64y$

Then you put in the x and y co-ordinates and determine whether is a maxima or minima as < or > 0.

But I don't have any numbers...so if anybody can help or see where I've gone wrong

Thanks

It seems you did not do the arithmetics.
As you have

$\displaystyle $\left\{ \begin{array}{l} \frac{{\partial f}}{{\partial x}} = {x^2} - 4{y^2} = 0\\ \\ \frac{{\partial f}}{{\partial y}} = {y^2} - 8xy + 15 = 0 \end{array} \right.$$

From the second one

$\displaystyle ${y^2} - 8xy + 15 = 0 \Rightarrow x = \frac{{{y^2} + 15}}{{8y}}$$

We go back to the first one

$\displaystyle ${x^2} - 4{y^2} = 0 \Rightarrow {\left( {\frac{{{y^2} + 15}}{{8y}}} \right)^2} - 4{y^2} = 0 \Rightarrow \frac{{{y^4} + 30{y^2} + 225 - 256{y^4}}}{{64{y^2}}} = 0$$

Which gives us

$\displaystyle $255{y^4} - 30{y^2} - 225 = 0 \Rightarrow 17{y^4} - 2{y^2} - 15 = 0 \Rightarrow (y - 1)(y + 1)(17{y^2} + 15) = 0$$

And exact values are at our hands

$\displaystyle ${x_1} = 2$$ and $\displaystyle ${y_1} = 1$$

$\displaystyle ${x_2} = - 2$$ and $\displaystyle ${y_2} = - 1$$
• Jan 15th 2011, 10:15 AM
ramdrop
Hm I tried subbing it in, but i re-arranged the first and went into the second, many thanks bud :)
• Jan 15th 2011, 02:11 PM
HallsofIvy
Quote:

Originally Posted by ramdrop
Hm I tried subbing it in, but i re-arranged the first and went into the second, many thanks bud :)

Perhaps you simply do not understand what is meant by "solve the equations". To solve
$x^2- 4y^2= 0$ and $y^2- 8xy+ 15= 0$
does NOT mean to solve one equation for x as a function of y (which is what you apparently did to get something like "(2y, -15)".

You can, as Pranas said, solve $x^2- 4y^2$ for x as a function of y: $x= \pm 2y$ and then put, first, x= 2y into the second equation: $y^2- 8xy+ 15= y^2- 8(2y)y+ 15= -15y^2+ 15$ which reduces to $15y^2= 15$ or $y= \pm 1$. Then, since x= 2y, x= 2(1)= 2 or x= 2(-1)= -2. Two solutions are (2, 1) and (-2, -1). Putting x= -2y into that equation instead, [tex]y^2- 8(-2y)y+ 15= 17y^2+ 15= 0 which has no real solutions.