# Examine whether function f(x,y)=x^2*y^2+1-4(x^2+y^2) has any stationary points?

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### 2 Answers

At the stationary points of any function f the derivative f' = 0.

Here f(x,y)=x^2*y^2+1-4(x^2+y^2)

So we need to find the partial derivatives with respect to x and y and equate them to 0.

df/dx = 2x*y^2 - 8x

2x*y^2 - 8x = 0

=> 2x(y^2 - 4) = 0

2x = 0 => x =0

y^2 = 4 => y = 2 and y = -2

df/dy = x^2*2y - 8y

x^2*2y - 8y = 0

2y(x^2 - 4) = 0

2y = 0 => y = 0

x^2 = 4 => x = 2 and x = -2

**The stationary points are (0, 0) , (2, 2) , (2 , -2), (-2 , 2) and (-2, -2)**

A function has stationary points if and only if the following equations are fulfiled:

df/dx = 0 and df/dy = 0

We'll calculate the partial derivative, with respect to x, assuming that y is a constant;

df/dx = 2y^2*x - 8x

df/dx = 2x(y^2 - 4)

We'll put df/dx = 0 => 2x(y^2 - 4) = 0

We'll set each factor as zero:

2x = 0

x = 0

y^2 - 4 = 0

y^2 = 4 => y1 = 2 and y2 = -2

We'll calculate the partial derivative df/dy, with respect to y, assuming that x is a constant:

df/dy = 2x^2*y - 8y

df/dy = 0

2y(x^2 - 4) = 0

2y = 0 => y3 = 0

x^2 - 4 = 0

x^2 = 4

x2 = 2 and x3 = -2

**We'll get 5 stationary points:(0,0) ; (2,2) ; (2,-2) ; (-2,-2) ; (-2,2).**