# find max and min of f(x,y) = (x - y)^2 xy

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### 1 Answer

You need to expand the square using the formula `(a-b)^2 = a^2 - 2ab + b^2` such that:

`f(x,y)= (x^2 - 2xy + y^2)xy => f(x,y) = x^3y - 2x^2y^2 + xy^3`

You need to evaluate the critical points, hence, you need to solve the equations `f_x=0` and `f_y=0` such that:

`f_x = 3x^2y - 4y^2x + y^3 => 3x^2y - 4y^2x + y^3 = 0`

`y(3x^2 - 4xy + y^2) = 0 => y = 0`

`3x^2 - 4xy + y^2 = 0 => 4xy = 3x^2 + y^2`

`f_y = x^3 - 4x^2y + 3xy^2 => x^3 - 4x^2y + 3xy^2 = 0`

`x(x^2 - 4xy + 3y^2) = 0 => x = 0`

Substituting `3x^2 + y^2` for `4xy` yields:

`x^2 -(3x^2 + y^2) + 3y^2 = 0`

`x^2 - 3x^2 + 3y^2 - y^2 = 0`

`-2x^2 + 2y^2 = 0 => -2(x^2 - y^2) = 0`

Converting the difference of squares into a product yields:

`x^2 - y^2 = (x-y)(x+y)`

`x^2 - y^2 = 0 => (x-y)(x+y) = 0 => x = +-y`

**Hence, evaluating the critical points of the given function yields that the only critical point is (0,0).**