Question

Locate the stationary points of x4+y4-2x2+4xy-2y2.

Answer 1

Answer: The stationary points of x⁴+y⁴-2x²+4xy-2y² are (0,0) (√2, -√2) and (-√2, √2)

Step-by-step explanation:

Let f(x,y) = x⁴+y⁴-2x²+4xy-2y²

We now differentiate f(x,y) with respect to x, treating y as a constant.

⇒ fₓ = 4x³-4x+4y

Similarly, differentiating f(x,y) with respect to y, treating x as a constant, we have,

fᵧ = 4y³+4x-4y

We know solve fₓ and fᵧ = 0.

fₓ = 0 ⇒ 4x³-4x+4y = 0

⇒ x³-x+y = 0.......(i)

⇒ x = x³+y

Again, if fᵧ = 0,

⇒ 4y³+4x-4y = 0

⇒ y³+x-y = 0........(ii)

Putting x = x³+y in the above equation, we have,

y³+x³+y-y = 0

⇒ y³+x³ = 0

⇒ y³ = -x³

Thus, we have, y = -x.

Putting y=-x in (i), we have,

x³-x-x = 0

⇒ x³-2x = 0

⇒ x (x²-2) = 0

⇒ x = 0 or x²-2 = 0

⇒ x = 0 or x² = 2

⇒ x = 0 or x = ±√2

Now, if x=0, y=0, if x=√2, then y= -√2 and if x=-√2, y= √2

Thus, the stationary points are (0,0) (√2, -√2) and (-√2, √2).