May 18th, 2018, 07:29 AM  #1 
Newbie Joined: May 2018 From: London Posts: 1 Thanks: 0  Classification of critical points
My question is to find critical points of: xy^2  x^2 +3x—y^4+ze^(z) and classify them. Please help, I have problems with finding the critical points, because the system of equation becomes big. 
May 18th, 2018, 08:23 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896  Three equations is really that big a system! You have $\displaystyle f(x, y, z)= xy^2 x^2+ 3x 4y^4+ ze^{z}$. Setting the partial derivatives to 0: $\displaystyle f_x= y^2 2x+ 3= 0$ $\displaystyle f_y= 2xy 16y^3= 0$ $\displaystyle f_z= e^{z} ze^{z}= 0$ The third equation involves only z: $\displaystyle e^{z}(1 z)= 0$. Since $\displaystyle e^{z}$ is never 0, z= 1. We now have $\displaystyle y^2 2x+ 3= 0$ and $\displaystyle 2xy 16y^3= y(2x 16y^2)= 0$. From the second, either y= 0 or $\displaystyle 2x=16y^2$. If y= 0, the first equation becomes 2x+ 3= 0 so x= 3/2. x= 3/2, y= 0, z= 1 is one solution. If $\displaystyle 2x= 16y^2$ then $\displaystyle x= 8y^2$ and the first equation becomes $\displaystyle y^2 2(8y^2)+ 3= 15y^2+ 3= 0$. $\displaystyle 15y^2= 3$ so $\displaystyle y^2= 1/5$and y is either $\displaystyle \frac{\sqrt{5}}{5}$ or $\displaystyle \frac{\sqrt{5}}{5}$. In the either case $\displaystyle x= 8y^2= \frac{8}{5}$. So $\displaystyle x= \frac{8}{5}$, $\displaystyle y= \frac{\sqrt{5}}{5}$, $\displaystyle z= 1$ is another solution and [math]$\displaystyle x= \frac{8}{5}$, $\displaystyle y= \frac{\sqrt{5}}{5}$, $\displaystyle z= 1$ is a third solution. 

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