有人会做这道高数题吗？

距离xoy最短就是|z|最小
令x=cost, y = sint ，这样(x,y,z)就在柱面上
带入平面方程
cost /3 + sint /4 + z/5 =1
z = 5 - 5cost/3 - 5sint /4

dz/dt = 5sint /3 +5 cost/4=0,得到t = -arctan(3/4)
此时sint = 3/5, cost = -4/5或者sint =-3/5, cost = 4/5
z= 5 -1 -1 =3 或者z=5 +1 +1 =7，舍去
所以最短距离为3

可以用拉氏乘数法：把柱面方程看做一约束
z = 5(1 - x/3 - y/4) + λ(x^2 + y^2 -1)
z'x = -5/3 + 2λx = 0 ==> λ = 5/(6x)
z'y = -5/4 + 2λy = 0 ==> λ = 5/(8y)
6x = 8y ==> y = 3x/4
代入：x^2 + y^2 = 1 ==> (3x/4)^2 + x^2 = 1
x = 4/5, y = 3/5
z = 5(1 - 4/15 - 3/20) = 5 - 4/3 - 3/4 = 35/12