Question

求解第三题，数学类的英文形式的题

第三大题就是如图，不是小题

Answer 1

(1) y = mx + c passes (2, 5), 5 = 2m + c, c = 5 - 2m, the line becomes y = mx + 5 - 2m
y = mx + 5 - 2m = 0, x = 2 - 5/m > 0, A(2 - 5/m, 0)
x = 0, y = 5 - 2m > 0, B(0, 5 - 2m)
S = 1/2)(2 - 5/m)(5 - 2m) = -2m - 25/(2m) + 10 ≥ 2√{(-2m)[-25/(2m)]} + 10 = 20, which is the min of the area of triangle AOB, when -2m = -25/(2m), m = -5/2 (ignore m = 5/2 > 0)

(2)
S = -2m - 25/(2m) + 10, m < 0
S' = -2 + 25/(2m²) = 0, m² = 25/4, m = -5/2 (drop m = 5/2 > 0)

(3)
y = mx + c passes (a, b), b = ma + c, c = b - ma, y = mx + b - ma
Similar as in (a), A(a - b/m, 0), B(0, b - ma)
S = (1/2)(a - b/m)(b - ma) = ab - a²m/2 - b²/(2m) ≥ 2√{(-a²m/2)[-b²/(2m)]} + ab
= ab + ab = 2ab, which is the min area (assuming (a, b) in the first quadrant).