⇒ 4h = 40 −10r ⇒ 2h = 20 − 5r Surface area of the cylinder = 2πr2+2πrh+πr(20−5r) = 20πr - 2πr2 = π (20r−3r2) So, we maximize (20r−3r2) for maximum surface area of a cylinder. Maximum value of a quadratic equation ax2 + bx + c = 0 arise for x =