Let two numbers be x and y, where x > 0, y > 0. Given, x + y = 9 ... (i) and z = x−y2 ... (ii) ⇒ z = (9−x)2 = x (81 + x2 - 18x) = x2−18x2 + 81x ∴
dz
dx
= 3x2 - 36x + 81 = 3 (x2 - 12x + 27)
dz
dx
= 0 ⇒ x2 - 9x - 3x + 27 = 0 ⇒ (x - 9) (x - 3) = 0 ⇒ x = 3 , x = 9 x = 3 [ x = 9 not possible] and y = 6 ∴ So, numbers are 3 and 6.