# How do you find the critical points(using partial derivative) for f( x , y )=x^3 + y^2 - 6x^2 + y -1?

Aug 20, 2015

Let Z = ${x}^{3}$ + ${y}^{2}$ - 6${x}^{2}$ + y -1

${f}_{x}$ = 3${x}^{2}$ - 12x
${f}_{y}$ = 2y + 1

${f}_{x}$ = 0 $\implies$ 3${x}^{2}$ - 12x = 0
x(3x - 12 ) = 0
x = 0
3x - 12 = 0
x = $\frac{12}{3}$ = 4

${f}_{y}$ = 0 $\implies$ 2y + 1 = 0

y = $- \frac{1}{2}$