Question

If x² + y² = 1, find the least value of f(x, y) = x² + y² + xy + 3.

Answer 1

Given x² + y² = 1

Put x = cosθ, y = sinθ

Now, f(θ) = cos²θ + sin²θ + sinθcosθ + 3

= 1 + 1/2 sin2θ + 3

= 1/2 sin2θ + 4

Max value = 1/2 .1 + 4 = 9/2

Min value = 1/2 .(–1) + 4 = 7/2

Hence, the least value of f(x, y) is 7/2