题目设x,y,z∈R+且x2(1/2)+y2+z=1,求xy+2xz的最大值.这是2010年北京大学自主招生试题,是一道含有三变元的条件最值问题,本题难度较大,很难找到解题入口,本文用主元法给出两种解法与大家分享.解法1依题意,设x=rcosθ,y=rsinθ,θ∈(0,π2),r∈(0,1),则x2(1/2)+y2+z=1为r+z=1,所以z=1-r.设w=xy+2xz,则w=r2sinθcosθ+2r(1-r)cosθ,把r当作主元θ当作 The subject set x, y, z∈R + and x2 (1/2) + y2 + z = 1, find the maximum value of xy + 2xz. This is the 2010 Peking University self-enrollment test is a condition containing the three variables Value problem, the problem is more difficult, it is difficult to find the entrance to solve the problem, this paper gives two ways to share with you with the method of principal component. Solution 1 according to the meaning, set x = rcosθ, y = rsinθ, θ∈ (0, ), r∈ (0,1), then x2 (1/2) + y2 + z = 1 is r + z = 1, so z = 1-r. Let w = xy + 2xz, then w = r2sinθcosθ + 2r (1-r) cos θ, taking r as the principal element θ