高中数学-6-业主生活–手机房天下问答

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emailsj 丨Lv 4

若实数x,y满足:3x^2-xy+3y^2=20.试求8x^2+23y^2的最大值.解设m=8x^2+23y^2,8x^2=m*(cost)^2,23y^2=m*(sint)^2.所以 xy=msin(2t)/(4√46).因为 3x^2-xy+3y^2=20 所以 20/m=(3/8)*[(1+cos(2t)]/2+(3/23)*[1-cos(2t)]/2-sin(2t)/(4√46)=93/(16*23)+45cos(2t)/(16*23)-sin(2t)/(4√46)≥93/(16*23)-√{[45/(16*23)]^2+[1/(4√46)]^2}=93/(16*23)-47/(16*23)=1/8故m≤160.

+1 2010-10-07 举报