題目 8. (10 points) Find the maximum value of the function f(x,y,z)=x+2y+3zf(x,y,z)=x+2y+3zf(x,y,z)=x+2y+3z on the curve of intersection of the plane x−y+z=1x-y+z=1x−y+z=1 and the cylinder x2+y2=1x^2+y^2=1x2+y2=1. 解答 解法一 由平面方程 z=1−x+yz=1-x+yz=1−x+y 消去 zzz，目標函數在圓柱交線上變為 f=3−2x+5y,x2+y2=1.f=3-2x+5y,\qquad x^2+y^2=1.f=3−2x+5y,x2+y2=1. 由柯西不等式，−2x+5y≤29-2x+5y\le\sqrt{29}−2x+5y≤29，等號可達。因此最大值為 3+29.\boxed{3+\sqrt{29}}.3+29.
