# Help defining critical pts and the sign of f(x,y)=(2x^2-y)*(x^2-y)

• Mar 7th 2013, 08:17 AM
lytwynk
Help defining critical pts and the sign of f(x,y)=(2x^2-y)*(x^2-y)
We were given this question in class as a exercise to do at home but I am confused about it...

Let F(x,y) =(2x2-y)(x2-y).
1. Discuss the sign of F at various points of the plane by appropriate consideration of the regions into which the plane is divided by the two parabolas y=x2 and y=2x2.
2. Discuss the critical points of the function.
3. Show that along every straight line through the origin, the values of F reach a minimum at (0,0), but that F neither has a minimum or a maximum at (0,0).

So far I don't really know how to do part 1. As far as 2 goes I have calculated (0,0) to be the only critical point... Is this correct? As for 3 I can show that (0,0) is a saddle point but I don't understand how to show that along every line through the origin the values of F reach a minimum at (0,0)...

Any help is greatly appreciated.
• Mar 7th 2013, 04:35 PM
HallsofIvy
Re: Help defining critical pts and the sign of f(x,y)=(2x^2-y)*(x^2-y)
As far as part 1 is concerned, you are given \$\displaystyle F(x,y)= (2x^2-y)(x^2-y)\$, do you not see that F(x,y)= 0 if and only if either \$\displaystyle 2x^2- y= 0\$ or \$\displaystyle x^2- y= 0\$? And those are equivalent to \$\displaystyle y= 2x^2\$ and \$\displaystyle y= x^23\$. Now since F is continuous and 0 only on those curves, F can change sign (go from positive to negative or vice-versa) only on those curves. So it is sufficient to test a single point in each of the regions bounded by those curves to see what sign F has in each.
Now \$\displaystyle F_x= 4x(x^2- y)+ 2x(2x^2- y)= 8x^3- 6xy= 2x(4x^2- 3y)= 0\$ which is 0 if x= 0 or if \$\displaystyle 4x^2= 3y\$, and \$\displaystyle F_y= -(x^2- y)- (2x^2- y)= -3x^2+ 2y= 0\$ so \$\displaystyle 2y= -3x^2\$. If x= 0, then 2y= 0 so y= 0. If \$\displaystyle 4x^2= 3y\$ then \$\displaystyle 2y= (8/3)x^4\$ so that \$\displaystyle 2y= (8/3)x^4= -3x^2\$, \$\displaystyle x^2((8/3)x^2+ 3)= 0\$ from which x= 0. Yes, (0, 0) is the only critical point.

"Any straight line" through the origin is of the form y= ax, for some a, or x= 0 (the vertical line through the origin). In the first case, y= ax, \$\displaystyle F(x, ax)= (2x^2- ax)(x^2- ax)= 2x^4- 3ax^3+ a^2x^2\$. The derivative of that is \$\displaystyle 8x^3- 9ax^2+ 4a^2x\$ which clearly has x= 0 as a 0. The second derivative is \$\displaystyle 24x^2- 9ax^2+ 4a\$ which is positive at x= 0 so F(x, ax) has a local minimum there. If x= 0, \$\displaystyle F(0, y)= (-y)(-y)= y^2\$ which clearly has a minimum at y= 0.
• Mar 8th 2013, 07:41 AM
lytwynk
Re: Help defining critical pts and the sign of f(x,y)=(2x^2-y)*(x^2-y)
This has been a huge help. Thank you very much.

I just have one final question about this function... The last part of the question asks to show that this function neither has a max or a min at this point, (0,0). Using the second derivative test,

Given F(x,y) = (2x2-y)(x2-y) = 2x4-3x2y+y2
Fx=8x3-6xy
Fy=-3x2+2y
A=Fxx=24x2-6y
B=Fxy=-6x
C=Fyy=2
D=B2-AC

At pt (0,0), A=0 and D=0 and from here no conclusion can be made about (0,0), it could be a max, a min or a saddle point... Is this good enough to answer this part of the question or am I missing something?