Question

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

Answer 1

Let Z = `x^3` + `y^2` - 6`x^2` + y -1

`f_x` = 3`x^2` - 12x
`f_y` = 2y + 1

`f_x` = 0 `=>` 3`x^2` - 12x = 0
x(3x - 12 ) = 0
x = 0
3x - 12 = 0
x = `12/3` = 4

`f_y` = 0 `=>` 2y + 1 = 0

y = `-1/2`