2 1，(ⅰ)AB = x，AC=y，AD= 1，∠BAC= 120，

Y=kx, x, y < 5, k= 1, y=x=2 (or √3), let BC=a, BD=b, CD=c,

Then a=b+c, cos 120 = (x? +y？ -a？ )/(2xy)=- 1/2，

arccos∠BAD+arccos∠CAD = 120，

cos∠BAD=(x？ + 1? -B? )/(2x)，

arcos∠CAD=(y？ + 1? -c？ )/(2y).

(ii) When y = x = ∠ 3, ∠ ACB = ∠ ABC = ∠ DAB = 30, BD=AD=, the total cost is the least.

△ABD area =( 1/4)√3, △ACD area =( 1/2)√3.

Answer these first.