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已知x,y∈[-1,1],则二元多项式P(x,y)=x^2+2xy+2y^2-y+1的取值范围是
人气:328 ℃ 时间:2020-06-23 14:50:59
解答
解由P(x,y)=x^2+2xy+2y^2-y+1=x^2+2xy+y^2+y²-y+1=(x+y)²+(y-1/2)²+3/4≥3/4即P(x,y)≥3/4当当y=-1,x=-1时P(x,y)有最大值=(x+y)²+(y-1/2)²+3/4=(-2)²+(-1-1/2)²+3/4=7...
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