Question

设x≥0.y≥0.x+2y＝0.5，求f（x）＝log1/2（8xy+4y^2+1）得最值

Answer 1

x+2y＝0.5 => y=(0.5-x)/2
x≥0.y≥0. => 0≤x≤0.5
f（x）＝log1/2（8xy+4y^2+1）定义域为8xy+4y^2+1>0
设 u=8xy+4y^2+1>0
则u=8x*(0.5-x)/2+4(0.5-x)^2/4+1
=4x(0.5-x)+(0.5-x)^2+1
=2x-4x^2+0.25-x+x^2+1
=1.25+x-3x^2
=5/4-3(1/6-x)^2+1/12
=4/3-3(1/6-x)^2
≤4/3
当0≤x≤0.5时，0.5/2=1/4>1/6，∴u的最小值为u(0.5)=4/3-1/3=1
∴u的取值范围为1≤u≤4/3
log1/2 u为减函数，∴最大值为log1/2 1=0，最小值为log1/2 4/3