ada yang bisa gak? perlu'

f(x) = 7 - √(25 - x^2) = 7 - (25 - x^2)^1/2
f'(x) = -1/2 (25 - x^2)^-1/2 . -2x = 0
=> x/(25 - x^2)^1/2 = 0
=> x/√(25 - x^2) = 0
=> x = 0
f(x) = 7 - √(25 - x^2)
f(0) = 7 - √(25 - 0^2) = 7 - √25 = 7 - 5 = 2
Jadi titik minimum​ = (0, 2) = (a, b)
a^2 + b^2 = 0^2 + 2^2 = 0 + 4 = 4