已知实数x、y满足x^2+y^2=1,则3x^2-x+4y^2的最小值

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已知实数x、y满足x^2+y^2=1,则3x^2-x+4y^2的最小值
已知实数x、y满足x^2+y^2=1,则3x^2-x+4y^2的最小值

已知实数x、y满足x^2+y^2=1,则3x^2-x+4y^2的最小值
y^2=1-x^2>=0
所以-1

x^2+y^2=1
y^2=1-x^2
3x^2-x+4(1-x^2)=3x^2-x+4-4x^2
=-x^2-x+4=-(x^2+x)+4
=-(x^2+x+1/4-1/4)+4
=-(x+1/2)^2+17/4
-1<=x<=1
当x=1 最小 =2

3x^2-x+4y^2=3x^2-x+(1-x^2)=2x^2-x+1=2(x^2-0.5x)+1
=2(x-0.5x+0.25^2-0.25^2)+1
=2(x-0.25)^2-0.0625*2+1
=2(x-0.25)^2+0.875
当x=0.25 存在最小值0.875
y^2=1-0.25^2=0.9375
y=根号0.9375 or y=-根号0.9375

可设x=cosa,y=sina,原式=-（cosa+1/2)^2+17/4.显然，当cosa=1时，原式有最小值2.

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