【数学I】例題3.2.9：条件付きの2変数関数の最大・最小1（One More）★★★

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$2x+y=1$を満たすとき，次の問いに答えよ．

(1) $x^2+y^2$の最小値を求めよ．

(2) $x\geqq 0,y\geqq 0$のとき，$x^2+y^2$の最大値を求めよ．

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(1) $2x+y=1$より，$y=1-2x\cdots(\mathrm{i})$

したがって，

\begin{align*}
x^2+y^2&=x^2+(1-2x)^2\\
&=5x^2-4x+1\\
&=5(x-\frac{2}{5})^2+\frac{1}{5}\cdots(\mathrm{ii})
\end{align*}

ゆえに，$x=\frac{2}{5}$で最小値$\frac{1}{5}$

このとき，(i)より，$y=1-2\cdot\frac{2}{5}=\frac{1}{5}$

よって，$x=\frac{2}{5},y=\frac{1}{5}$のとき，最小値$\frac{1}{5}$

(2) $y\geqq 0$であるから，(i)より，$1-2x\geqq 0$

$x\geqq 0$との共通範囲は，$0\leqq x\leqq\frac{1}{2}\cdots(\mathrm{iii})$

(ii)より，(iii)において，$x^2+y^2$は$x=0$で最大値$1$

このとき，(i)より，$y=1-2\cdot 0=1$

よって，$x=0,y=1$のとき，最大値$1$

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$x+3y=6$を満たすとき，次の問いに答えよ．

(1) $x^2+y^2$の最小値を求めよ．

(2) $x\geqq 0,y\geqq 0$のとき，$x^2+y^2$の最大値を求めよ．

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(1) $x+3y=6$より，$y=2-\frac{x}{3}\cdots(\mathrm{i})$

したがって，

\begin{align*}
x^2+y^2&=x^2+(2-\frac{x}{3})^2\\
&=x^2+(4-\frac{4x}{3}+\frac{x^2}{9})\\
&=\frac{10}{9}x^2-\frac{4}{3}x+4\\
&=\frac{10}{9}(x-\frac{3}{5})^2+\frac{18}{5}\cdots(\mathrm{ii})
\end{align*}

ゆえに，$x=\frac{3}{5}$で最小値$\frac{18}{5}$

このとき，(i)より，$y=2-{\frac{3}{5}}\div{3}=\frac{9}{5}$

よって，$x=\frac{3}{5},y=\frac{9}{5}$のとき，最小値$\frac{18}{5}$

(2) $y\geqq 0$であるから，(i)より，$2-\frac{x}{3}\geqq 0$

$x\geqq 0$との共通範囲は，$0\leqq x\leqq 6\cdots(\mathrm{iii})$

(ii)より，(iii)において，$x^2+y^2$は$x=6$で最大値$36$

このとき，(i)より，$y=2-\frac{6}{3}=0$

よって，$x=6,y=0$のとき，最大値$36$

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% 例題I3.2.9の側注：条件付きの2変数関数の最大・最小1（One More）★★★
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% 側注1
$y$を代入して，文字を消去する．

% 側注2
$x\geqq 0,y\geqq 0$から，$x$の値の範囲を求める．

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% 問題I3.2.9の側注：条件付きの2変数関数の最大・最小1（One More）★★★
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% 側注1
$y$を代入して，文字を消去する．なお，$x=6-3y$として，$x$を代入して，文字を消去してもよい．

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% 例題I3.2.9の考え方・補足など：条件付きの2変数関数の最大・最小1（One More）★★★
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変数が少ない方が扱いやすいため，文字の消去を考える．

(1) 条件から$2x+y=1$は，$y=1-2x$と変形できるため，$x^2+y^2$に代入すると，$x^2+y^2=x^2+(1-2x)^2$となる．これは，文字が消去されて1変数の式であるから，1変数の2次関数の最小値を求める問題とすることができる．

(2) $x\geqq 0,y\geqq 0$の条件より，(1)の2次関数における文字の値の範囲を求めてから，最大値を考える．

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変数を少なくするために，文字の消去を考える．