【数学I】例題3.3.31：条件付きの2変数関数の最大・最小2（One More）★★★

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% 例題I3.3.31：条件付きの2変数関数の最大・最小2（One More）★★★
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実数$x,y$が$x^2+y^2=1$を満たすとき，$\frac{1}{2}x+y^2$の最大値，最小値と，そのときの$x,y$の値を求めよ．

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% 例題I3.3.31の解答（One More）★★★
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$x^2+y^2=1$より，$y^2=1-x^2\cdots(\mathrm{i})$

$x,y$は実数であるから$y^2\geqq 0$，すなわち，$1-x^2\geqq 0$

したがって，$x^2-1\leqq 0$であるから，$(x+1)(x-1)\leqq 0$

ゆえに，$-1\leqq x\leqq 1$

$\frac{1}{2}x+y^2$に(i)を代入すると，

\begin{align*}
\frac{1}{2}x+y^2&=\frac{1}{2}x+(1-x^2)\\
&=-x^2+\frac{1}{2}x+1\\
&=-(x-\frac{1}{4})^2+\frac{17}{16}
\end{align*}

グラフは右の図のようになる．

したがって，$x=\frac{1}{4}$のとき，最大値$\frac{17}{16}$，$x=-1$のとき，最小値$-\frac{1}{2}$

$x=\frac{1}{4}$のとき，$y^2=1-(\frac{1}{4})^2=\frac{15}{16}$より，$y=\pm\frac{\sqrt{15}}{4}$

$x=-1$のとき，$y^2=1-(-1)^2=0$より，$y=0$

よって，$\frac{1}{2}x+y^2$は，

\begin{align*}
&x=\frac{1}{4},y=\pm\frac{\sqrt{15}}{4}\text{のとき，}&&\text{最大値}\frac{17}{16}\\
&x=-1,y=0\text{のとき，}&&\text{最小値}-\frac{1}{2}
\end{align*}

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% 問題I3.3.31：条件付きの2変数関数の最大・最小2（One More）★★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

実数$x,y$が$x^2+y^2=1$を満たすとき，$x+y^2$の最大値，最小値と，そのときの$x,y$の値を求めよ．

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% 問題I3.3.31の解答（One More）★★★
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$x^2+y^2=1$より，$y^2=1-x^2\cdots(\mathrm{i})$

$x,y$は実数であるから$y^2\geqq 0$，すなわち，$1-x^2\geqq 0$

したがって，$x^2-1\leqq 0$であるから，$(x+1)(x-1)\leqq 0$

ゆえに，$-1\leqq x\leqq 1$

$x+y^2$に(i)を代入すると，

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

グラフは右の図のようになる．

したがって，$x=\frac{1}{2}$のとき，最大値$\frac{5}{4}$，$x=-1$のとき，最小値$-1$

$x=\frac{1}{2}$のとき，$y^2=1-(\frac{1}{2})^2=\frac{3}{4}$より，$y=\pm\frac{\sqrt{3}}{2}$

$x=-1$のとき，$y^2=1-(-1)^2=0$より，$y=0$

よって，$x+y^2$は

\begin{align*}
&x=\frac{1}{2},y=\pm\frac{\sqrt{3}}{2}\text{のとき，}&&\text{最大値}\frac{5}{4}\\
&x=-1,y=0\text{のとき，}&&\text{最小値}-1
\end{align*}

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% 例題I3.3.31の側注：条件付きの2変数関数の最大・最小2（One More）★★★
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% 側注1
$y$が実数のとき，$y^2\geqq 0$であることを用いて，$x$の値の範囲を求める．

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% 問題I3.3.31の側注：条件付きの2変数関数の最大・最小2（One More）★★★
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% 側注1
$y$が実数のとき，$y^2\geqq 0$であることを用いて，$x$の値の範囲を求める．

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% 例題I3.3.31の考え方・補足など：条件付きの2変数関数の最大・最小2（One More）★★★
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% 考え方
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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．ここでは，$x$よりも$y$の方が消去しやすいので，$x^2+y^2=1$を$y^2=1-x^2$と変形し，$\frac{1}{2}x+y^2$に代入する．すると，2次関数の最大値，最小値を求める問題とすることができる．このとき，$x$の範囲に注意すること．$(\text{実数})^2\geqq 0$，すなわち，$y^2\geqq 0$から，$x$の範囲を求めることができる．

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実数の条件$y^2\geqq 0$を忘れないように注意すること．