若3x2-xy+3y2=20,则8x2+23y2的最大值
人气：281 ℃ 时间：2019-12-24 10:37:26
解答
俊朗精英团队为你
20=3x^2+3y^2-xy=3x^2+3y^2-4*(1/4xy)>=3x^2+3y^2-4(x^2+1/16y^2)*1/2
=x^2+3y^2-1/8y^2=1/8(8x^2+23y^2)
8x^2+23y^2>=160
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