# Critical point question

• Mar 12th 2010, 05:17 AM
anderson
Critical point question
hi everyone

having problems finding critical points for this two equation to fined the maximum,minimum or saddle points.

a) \$\displaystyle x^4+y^4-4xy+1\$
\$\displaystyle f_x=4x^3-4y\$, \$\displaystyle f_y=4y^3-4x\$

b) f(x,y)=\$\displaystyle x^4-2x^2y+2y\$
\$\displaystyle f_x=4x^3-4xy\$,\$\displaystyle f_y=2-2x^2\$

need help to find the critical points for this two equations,really hope someone can help me. really appreciate all your help & support.
• Mar 12th 2010, 05:29 AM
Sputnik
How far have you come in your attempts? Where do critical points for a given function occur?
• Mar 12th 2010, 05:44 AM
anderson

a) \$\displaystyle x^3-y=0\$,\$\displaystyle y^3-x=0\$

\$\displaystyle x^3=y\$
\$\displaystyle y^3=x\$
im stuck here, unable to find the critical point
b)\$\displaystyle 4x^3-4xy=0\$
\$\displaystyle 4x^3=4xy\$
\$\displaystyle x^2=y\$

\$\displaystyle 2-2x^2=0\$
\$\displaystyle 2=2x^2\$
x=1 & -1

critical point, = (1,1) & (-1,1) ..is this right???

realyl hope someone can help, appreciate all help & support.
• Mar 12th 2010, 06:16 AM
Sputnik
a: Remember that the two derivatives form a system of equations:
\$\displaystyle x^3-y = 0\$
\$\displaystyle y^3-x = 0\$

imply that

\$\displaystyle x^9=x\$ or \$\displaystyle x(x^8-1) = 0\$ (see why?)

What values for critical points does this give you?

b: Seems right to me, so far, however, you are being a little hasty. It's quite right that
\$\displaystyle 4x^3-4xy = 0\$ implies that
\$\displaystyle x^2 = y\$
but it also implies that
\$\displaystyle x(x^2-y) = 0\$. See what you missed?
• Mar 12th 2010, 06:45 AM
anderson