【数学I】例題3.2.7：最小値の最大値（One More）★★★

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% 例題I3.2.7：最小値の最大値（One More）★★★
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$x$の2次関数$y=x^2-6ax+8a^2+2a+3$の最小値を$m$とする．このとき，次の問いに答えよ．ただし，$a$は定数とする．

(1) 最小値$m$を$a$を用いて表せ．

(2) $a$の値が$0\leqq a\leqq 3$で変化するとき，$m$の最大値を求めよ．

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(1)

\begin{align*}
y=&x^2-6ax+8a^2+2a+3\\
=&\{(x-3a)^2-(3a)^2\}+8a^2+2a+3\\
=&(x-3a)^2-a^2+2a+3
\end{align*}

グラフは右の図のようになる．

よって，$y$は$x=3a$で最小値$m=-a^2+2a+3$

(2)

\begin{align*}
m&=-a^2+2a+3\\
&=-(a^2-2a)+3\\
&=-\{(a-1)^2-1^2\}+3\\
&=-(a-1)^2+4
\end{align*}

グラフは右の図のようになる．

よって，$0\leqq a\leqq 3$の範囲において，$a$の関数$m$は，$a=1$で最大値$4$

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% 問題I3.2.7：最小値の最大値（One More）★★★
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$x$の2次関数$y=-x^2+4ax-5a^2+3a+6$の最大値を$M$とする．このとき，次の問いに答えよ．ただし，$a$は定数とする．

(1) 最大値$M$を$a$を用いて表せ．

(2) $a$の値が$-2\leqq a\leqq 3$で変化するとき，$M$の最小値を求めよ．

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(1)

\begin{align*}
y=&-x^2+4ax-5a^2+3a+6\\
=&-\{(x-2a)^2-(2a)^2\}-5a^2+3a+6\\
=&-(x-2a)^2-a^2+3a+6
\end{align*}

グラフは右の図のようになる．

よって，$y$は$x=2a$で最大値$M=-a^2+3a+6$

(2)

\begin{align*}
M&=-a^2+3a+6\\
&=-(a^2-3a)+6\\
&=-\{(a-\frac{3}{2})^2-(\frac{3}{2})^2\}+6\\
&=-(a-\frac{3}{2})^2+\frac{33}{4}
\end{align*}

グラフは右の図のようになる．

よって，$-2\leqq a\leqq 3$の範囲において，$a$の関数$M$は，$a=-2$で最小値$-4$

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% 例題I3.2.7の側注：最小値の最大値（One More）★★★
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% 側注1
平方完成する．

% 側注2
$m$は$a$の2次式で表される．

% 側注3
$x$軸，$y$軸ではなく，$a$軸，$m$軸であることに注意すること．

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% 問題I3.2.7の側注：最小値の最大値（One More）★★★
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% 側注1
平方完成する．

% 側注2
$M$は$a$の2次式で表される．

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% 例題I3.2.7の考え方・補足など：最小値の最大値（One More）★★★
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% 考え方
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(2) (1)で求めた最小値$m$を$a$の関数と見ると，定義域における最大・最小の問題となる．$a$を変数として，関数$m=-a^2+2a+3$の最大値を求める．

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「最小値の最大値」と聞いても，具体的にイメージすることができないかもしれない．$y=x^2-6ax+8a^2+2a+3$のグラフは，$a$の値を1つ定めると，その配置がただ1つに定まる．頂点の$y$座標が最小値$m$であり，$m$は$m=-a^2+2a+3$と表されるから，最小値は$a$の関数であり，グラフは上に凸の放物線となる．(2)は$0\leqq a\leqq 3$におけるそのグラフの最大値を求めている．イメージとしては，右の図のようになる．