Question

高等数学应用题，最好有过程

Answer 1

设长宽高分别为 x, y, z, 构造拉格朗日函数
F = 2(xy+yz+zx)+k(xyz-2)
F'<x> = 2(y+z)+kyz = 0 (1)
F'<y> = 2(x+z)+kxz = 0 (2)
F'<z> = 2(y+x)+kxy = 0 (3)
F'<k> = xyz-2 = 0 (4)
(1)x 得 2x(y+z)+kxyz = 0, （4） 代入，得 x(y+z) = -k
类似地 (2)y 得 y(x+z) = -k, (3)z 得 z(x+y) = -k
故 xy+xz = xy+yz = xz+yz
则 x = y = z,
又 xyz = 2， 则 x = y = z = 2(1/3) m,
即正方体时 材料最省。

Answer 2

设长宽高分别为 x, y, z, 构造拉格朗日函数
F = 2(xy+yz+zx)+k(xyz-2)
F'<x> = 2(y+z)+kyz = 0 (1)
F'<y> = 2(x+z)+kxz = 0 (2)
F'<z> = 2(y+x)+kxy = 0 (3)
F'<k> = xyz-2 = 0 (4)
(1)x 得 2x(y+z)+kxyz = 0, （4） 代入，得 x(y+z) = -k
类似地 (2)y 得 y(x+z) = -k, (3)z 得 z(x+y) = -k
故 xy+xz = xy+yz = xz+yz
则 x = y = z,
又 xyz = 2， 则 x = y = z = 2(1/3) m,
即正方体时 材料最省。