partial derivatives of rel min and max values

thelazyman

Junior Member
Joined
Jan 14, 2006
Messages
58
Hi, there are 2 systems of equations here and Ineed help in simplfying them

The first one is:

f1 = 3x^2 - 6y = 0
f2 = 3y^2 -6x = 0

I obtained

y = x^2/2 when plugging into y i got 3(x^2/2)^2 - 6x = 0

I then run into a problem.

The notes say the next step are

x(x^3-2^3). I do not understand where they got the 8 because when I solve it i got 12x^4 - 6x and I do not understand how that simplifys to x(x^3-2^3)

Someone please help.

thanks
 
thelazyman said:
Hi, there are 2 systems of equations here and Ineed help in simplfying them

The first one is:

f1 = 3x^2 - 6y = 0
f2 = 3y^2 -6x = 0

I obtained

y = x^2/2 when plugging into y i got 3(x^2/2)^2 - 6x = 0

I then run into a problem.

The notes say the next step are

x(x^3-2^3). I do not understand where they got the 8 because when I solve it i got 12x^4 - 6x and I do not understand how that simplifys to x(x^3-2^3)

Someone please help.

thanks

\(\displaystyle \L 3(\frac{x^2}{2})^2 -6x = 0 \,\, \Rightarrow \\
\frac{4}{3} \[ 3(\frac{x^2}{2})^2 -6x \] = \frac{4}{3} \[ 0 \] \,\, \Rightarrow \\
x^4 - 8x = 0 \,\, \Rightarrow \,\, x(x^3-2^3)=0\)
 
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