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40X+90Y+20XY
人气:246 ℃ 时间:2020-10-01 17:43:50
解答
不妨设xy≥0,40x+90y+20xy≤32000≥4x+9y+2xy-320≥12√(xy)+2xy-320=2xy+12√(xy)-320=2{[√(xy)]^2+6√(xy)+9}-338=2[√(xy)+3]^2-3382[√(xy)+3]^2-338≤0[√(xy)+3]^2-169≤0-13≤√(xy)+3≤13-16≤√(xy)≤100...
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