If x^2+y^2=1, what is the largest possible value of |x|+|y|?

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.

a.

step-by-step explanation:

step 1 distribute 2 to every term in the parentheses

step 2 add 10 on both sides of the equal sign