If x² + y² = 1, then y = ±√(1 - x²).
Let f(x) = |x| + |±√(1 - x²)| = |x| + √(1 - x²).
If x < 0, we have |x| = -x ; otherwise, if x ≥ 0, then |x| = x.
• Case 1: suppose x < 0. Then
f(x) = -x + √(1 - x²)
f'(x) = -1 - x/√(1 - x²) = 0 → x = -1/√2 → y = ±1/√2
• Case 2: suppose x ≥ 0. Then
f(x) = x + √(1 - x²)
f'(x) = 1 - x/√(1 - x²) = 0 → x = 1/√2 → y = ±1/√2
In either case, |x| = |y| = 1/√2, so the maximum value of their sum is 2/√2 = √2.
