x,y为何值时,多项式4x²+9y²-4x+12y-1有最小值,并求出最小值?
题目
x,y为何值时,多项式4x²+9y²-4x+12y-1有最小值,并求出最小值?
4x²+9y²-4x+12y-1
=(4x²-4x+1)+(9y²+12y+4)-6
=(2x-1)²+(3y+2)²-6
由于(2x-1)²、(3y+2)²≥0,所以原多项式的最小值当(2x-1)²=0且(3y+2)²=0时取得,为-6,
解得此时的x=1/2,y=-2/3,
因此,当x=1/2,y=-2/3时,多项式4x²+9y²-4x+12y-1有最小值为-6.
这答案为什么解得x=1/2,y=2/3
答案
答:4x²+9y²-4x+12y-1=(4x²-4x+1)+(9y²+12y+4)-6=(2x-1)²+(3y+2)²-6因为:(2x-1)²>=0,(3y+2)²>=0所以:4x²+9y²-4x+12y-1=(2x-1)²+(3y+2)²-6>=0+0-6=-6...
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