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已知x^2+2y^2=1.求2x+5y^2的最小值
人气:133 ℃ 时间:2020-05-08 19:52:09
解答
x^2+2y^2=1
|x|<=1
y^2=(1-x^2)/2
2x+5y^2
=2x+(1-x^2)(5/2)
=-(5/2)x^2+2x+(5/2)=f(x)
|x|<=1
f(-1)=min=-2
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