# FInding Critical Point Using Partial Derivative

• Mar 21st 2009, 09:16 PM
Hakhengkim
FInding Critical Point Using Partial Derivative
I want to confirm if I am right,

The critical points that i got from this equation

C = x4 - 8xy + 2y2 - 3

is (0,0)
• Mar 21st 2009, 09:47 PM
Spec
(0,0) is not the only critical point.
• Mar 21st 2009, 09:49 PM
Hakhengkim
Quote:

Originally Posted by Spec
(0,0) is not the only critical point.

Really what are the other ones (Worried). By any chance can you show me?
• Mar 21st 2009, 09:50 PM
Chris L T521
Quote:

Originally Posted by Hakhengkim
I want to confirm if I am right,

The critical points that i got from this equation

C = x4 - 8xy + 2y2 - 3

is (0,0)

The function is $\displaystyle f\!\left(x,y\right)=x^4-8xy+2y^2-3$

Thus, $\displaystyle \frac{\partial f}{\partial x}=4x^3-8y$ and $\displaystyle \frac{\partial f}{\partial y}=4y-8x$

Now to find the critical points, solve $\displaystyle \frac{\partial f}{\partial x}=0$ and $\displaystyle \frac{\partial f}{\partial y}=0$

This would mean that $\displaystyle 4x^3-8y=0$ and $\displaystyle 4y-8x=0$.

Solving the second equation, we have $\displaystyle 4y=8x$

Substituting this into the first equation, we have $\displaystyle 4x^3-2(8x)=0\implies 4x^3-16x=0\implies 4x(x^2-4)=0$

Thus, $\displaystyle x=0,~x=2$ or $\displaystyle x=-2$.

As a result, when:

$\displaystyle x=0\,:\, 4y=8(0)\implies y=0$

$\displaystyle x=2\,:\, 4y=8(2)\implies y=4$

$\displaystyle x=-2\,:\, 4y=8(-2)\implies y=-4$

Thus, your critical points should be $\displaystyle \left(0,0\right)$, $\displaystyle \left(2,4\right)$ and $\displaystyle \left(-2,-4\right)$

Does this make sense?
• Mar 21st 2009, 10:00 PM
Hakhengkim
Quote:

Originally Posted by Chris L T521
The function is $\displaystyle f\!\left(x,y\right)=x^4-8xy+2y^2-3$

Thus, $\displaystyle \frac{\partial f}{\partial x}=4x^3-8y$ and $\displaystyle \frac{\partial f}{\partial y}=4y-8x$

Now to find the critical points, solve $\displaystyle \frac{\partial f}{\partial x}=0$ and $\displaystyle \frac{\partial f}{\partial y}=0$

This would mean that $\displaystyle 4x^3-8y=0$ and $\displaystyle 4y-8x=0$.

Solving the second equation, we have $\displaystyle 4y=8x$

Substituting this into the first equation, we have $\displaystyle 4x^3-2(8x)=0\implies 4x^3-16x=0\implies 4x(x^2-4)=0$

Thus, $\displaystyle x=0,~x=2$ or $\displaystyle x=-2$.

As a result, when:

$\displaystyle x=0\,:\, 4y=8(0)\implies y=0$

$\displaystyle x=2\,:\, 4y=8(2)\implies y=4$

$\displaystyle x=-2\,:\, 4y=8(-2)\implies y=-4$

Thus, your critical points should be $\displaystyle \left(0,0\right)$, $\displaystyle \left(2,4\right)$ and $\displaystyle \left(-2,-4\right)$

Does this make sense?

Ahhh ic thank you totally forgot about the +4y on the second partial derivative.