Find the extrema of f(x,y)=xy−y−x+1 on the disk x^2+y^2 ≤2?

∇f = (y - 1)i + (x - 1)j

This is zero if (x, y) = (1, 1). This is the only critical point on the disk (its boundary as it were). The value of f there is f(1, 1) = 0.

To determine any other extrema on the boundary, it's useful to write x = √(2)cos(t) and y = √(2)sin(t). You can write y = √(2 - x²) and then y = -√(2 - x²). But then you need to consider both pieces of the curve. The parametric representation lets you kill all birds with the one stone.

f(x, y) = 2sin(t)cos(t) - √(2)sin(t) - √(2)cos(t) + 1 on the circle.

With a little simplification, we get

df/dt = (sin(t) - cos(t))(√(2) - 2sin(t) - 2cos(t)).

The first factor is zero if t = π/4 or 5π/4. The second factor is zero if t = 7π/12 or 23π/12. These correspond to

t = π/4, (x, y) = (1, 1), f(1, 1) = 0

t = 5π/4, (x, y) = (-1, -1), f(-1, -1) = 4

t = 7π/12, (x, y) = (½ - √(3)/2, ½ + √(3)/2), f(x, y) = -1/2

t = 23π/12, (x, y) = (½ + √(3)/2, ½ - √(3)/2), f(x, y) = -1/2.

The absolute max and min are 4 and -1/2, respectively.

A fast way to find the two roots proportional to π/12 is to write sin(t) + cos(t) = √(2)sin(t + π/4).

****** You sent me a message, but don't allow people to respond to your messages?!

No, it wouldn't be x = 2cos(t) and y = 2sin(t). Those would give x² + y² = 2² = 4. The problem says the boundary is x² + y² = 2.