Question

f(x,y,z)=xyz,x²+2y²+3z²=6，求最大最小值

Answer 1

依三元均值不等式得
6=x²+2y²+3z²
≥3·(x²·2y²·3z²)^(1/3)
→6(xyz)²≤8
→-2/√3≤xyz≤2/√3
→-2√3/3≤f(x,y,z)≤2√3/3.
故所求最大值f(x,y,z)|max=2√3/3;
所求最小值为f(x,y,z)|min=-2√3/3。