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[函数] 求a+b最小值

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realnumber posted 2019-5-20 11:20 |Read mode
已知$2bt^2+3at-2b-3\le 0$,对任意$t\in [-\sqrt{2},\sqrt{2}]$恒成立,则a+b的最小值是_______.

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kuing posted 2019-5-20 12:57
看:forum.php?mod=viewthread&tid=3738,照搬即可。

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original poster realnumber posted 2019-5-20 13:55
按2楼提供办法得到t=-0.5,得到a+b ≥-2,此时a=-0.8,b=-1.2
我得想想为什么可以......

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敬畏数学 posted 2019-5-20 15:05
$ a+b\geqslant -2,-\frac{12}{5} (t+\frac{1}{2})^2\leqslant 0$

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isee posted 2019-5-21 20:15
回复 3# realnumber


    见到2#我就想起来了,不过,这之前真是忘记了。

   

    按所求的结合,最直接的想法就是让$a<0,b<0$(等于零,另外讨论),如果运气好的话,就是答案,(否则还是需要讨论。)

    动笔算了下,满足的条件即

\begin{align*}
a&<0,\\
b&<0,\\
-3\sqrt 2 a+2b-3&\leqslant 0,\\
9a^2+16b^2+24b&\leqslant 0
\end{align*}
求$(a+b)$的最大值,而恰好是与椭圆切的时候:$-2$.

这题的数据实在是太特别了。

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敬畏数学 posted 2019-5-21 23:49
常规方法也可以的。

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