Question

Find the max or min values of f(x, y) = x² + y² – xy, where x² + 4y² = 4

Answer 1

We have x² + 4y² = 4

⇒  x²/4 + y²/1 = 1

Put x = 2cos²θ and y = sinθ

Then f(x, y) = x² + y² – xy

= 4cos²θ + sin²θ – sin2θ

= 2(2cos²θ) + 1/2(2sin²θ) – sin2θ

= 2(1 + cos2θ) + 1/2(1 – cos2θ) – sin2θ

= 3/4cos2θ – sin2θ + 5/2

Max value = √(9/4)  + 1 + 5/2)  = (5 + √13)/2

Min value = – √(9/4)  + 1 + 5/2 = (5 – √13)/2