Finding extreme values by partial differentiation.

[toreil]

Homework Statement

I sat my calculus exam at the end of june and of the questions on the paper required us to find the extreme values of the following equation:

$g\left(x,y\right) = \left(x-1\right)\left(y-1\right)\left(x^{2} + y^{2} -2\right)$

The Attempt at a Solution

So i get:

$\frac{\partial g}{\partial x} = 0 = \left(y-1\right)\left(3x^{2}-2x+y^{2} -2\right)$

$\frac{\partial g}{\partial y} = 0 = \left(x-1\right)\left(3y^{2}-2y+x^{2} -2\right)$

From this I get 3 critical points $\left(1,1\right), \left(1,-1\right), \left(-1,1\right)$.

However in the answer sheet there are three more, but for the life of me I can't figure out how to get them. Is there a special trick that I'm missing out on? Any help would be much appreciated!

Final 3 points are included below for those who want to know them

$\left(-\frac{1}{2},-\frac{1}{2}\right)$
$\left(\frac{1}{2}-\frac{1}{\sqrt[]{2}},\frac{1}{2}+\frac{1}{\sqrt[]{2}}\right)$
$\left(\frac{1}{2}+\frac{1}{\sqrt[]{2}},\frac{1}{2}-\frac{1}{\sqrt[]{2}}\right)$

 

[ehild]

The other solutions come from the system of equations

3x²-2x+y^2-2=0 and 3y²-2y+x^2-2=0.

ehild

 

[toreil]

That's what I assumed to be the case, but unless I am doing something wrong I end up with

$\left(x-\frac{1}{2}\right)^2 - \left(y-\frac{1}{2}\right)^2 = 0$

So the point is $\left(\frac{1}{2},\frac{1}{2}\right)$

Thanks for the quick reply though!

 

[SammyS]

That's what I assumed to be the case, but unless I am doing something wrong I end up with

$\left(x-\frac{1}{2}\right)^2 - \left(y-\frac{1}{2}\right)^2 = 0$

So the point is $\left(\frac{1}{2},\frac{1}{2}\right)$

Thanks for the quick reply though!

How did you solve $\displaystyle \left(x-\frac{1}{2}\right)^2 - \left(y-\frac{1}{2}\right)^2 = 0\,?$ There are more solutions than you gave.

$\displaystyle \left(x-\frac{1}{2}\right)^2 = \left(y-\frac{1}{2}\right)^2$

taking the square root of both sides gives:
$\displaystyle \left(x-\frac{1}{2}\right) = \pm\left(y-\frac{1}{2}\right)$
​

Can you solve that. The solution set is the union of two lines.

 

[toreil]

Wow yeah thanks, need to brush up on my more basic maths again I think. Thanks for the help from both of you! That question has been bugging me for a few weeks now =P.