You need both partials to be zero. That is,
#1: 3y^3 + 2xy = 0
#2: 3(3x+1)y^2 + x^2 = 0
From #1, we get x = -3y^2/2
Plug that into #2 and you get y^2 = 4/15
So, z has stationary points at (-2/5, ±2/√15)
find the stationary point of the function
z=4x^2+10xy+4y^2-x^2y^2
and also determine their nature
plz show me step i don't know it
1 answer