【数学I】例題3.3.32：条件なし2変数関数（One More）★★★

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% 例題I3.3.32：条件なし2変数関数（One More）★★★
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次の関数の最小値と，そのときの$x,y$の値を求めよ．

(1) $x,y$の関数$P=x^2+2y^2+6x-4y+3$の最小値を求めよ．

(2) $x,y$の関数$Q=x^2-2xy+2y^2+4x-3y+8$の最小値を求めよ．

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% 例題I3.3.32の解答（One More）★★★
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(1)

\begin{align*}
P&=x^2+6x+2y^2-4y+3\\
&=(x+3)^2-3^2+2y^2-4y+3\\
&=(x+3)^2+2(y-1)^2-2\cdot 1^2-6\\
&=(x+3)^2+2(y-1)^2-8
\end{align*}

$x,y$は実数であるから，

\[
(x+3)^2\geqq 0,(y-1)^2\geqq 0
\]

したがって，$P$は$x+3=0,y-1=0$のとき最小となる．

よって，$x=-3,y=1$のとき，最小値$-8$

(2)

\begin{align*}
Q&=x^2-2xy+2y^2+4x-3y+8\\
&=x^2-2(y-2)x+2y^2-3y+8\\
&=\{x-(y-2)\}^2-(y-2)^2+2y^2-3y+8\\
&=(x-y+2)^2+y^2+y+4\\
&=(x-y+2)^2+(y+\frac{1}{2})^2-(\frac{1}{2})^2+4\\
&=(x-y+2)^2+(y+\frac{1}{2})^2+\frac{15}{4}
\end{align*}

$x,y$は実数であるから，

\[
(x-y+2)^2\geqq 0,(y+\frac{1}{2})^2\geqq 0
\]

したがって，$Q$は$x-y+2=0,y+\frac{1}{2}=0$のとき最小となる．

$x-y+2=0,y+\frac{1}{2}=0$を解くと，$x=-\frac{5}{2},y=-\frac{1}{2}$

よって，$x=-\frac{5}{2},y=-\frac{1}{2}$のとき，最小値$\frac{15}{4}$

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% 問題I3.3.32：条件なし2変数関数（One More）★★★
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次の関数の最小値と，そのときの$x,y$の値を求めよ．

(1) $x,y$の関数$P=x^2+3y^2-4x+2y+5$の最小値を求めよ．

(2) $x,y$の関数$Q=x^2+4xy+5y^2-4x-4y+7$の最小値を求めよ．

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(1)

\begin{align*}
P&=x^2-4x+3y^2+2y+5\\
&=(x-2)^2-2^2+3y^2+2y+5\\
&=(x-2)^2+3(y+\frac{1}{3})^2-3\cdot(\frac{1}{3})^2+1\\
&=(x-2)^2+3(y+\frac{1}{3})^2+\frac{2}{3}
\end{align*}

$x,y$は実数であるから，$(x-2)^2\geqq 0,(y+\frac{1}{3})^2\geqq 0$

したがって，$P$は$x-2=0,y+\frac{1}{3}=0$のとき最小となる．

よって，$x=2,y=-\frac{1}{3}$のとき，最小値$\frac{2}{3}$

(2)

\begin{align*}
Q&=x^2+4xy+5y^2-4x-4y+7\\
&=x^2+2(2y-2)x+5y^2-4y+7\\
&=\{x+(2y-2)\}^2-(2y-2)^2+5y^2-4y+7\\
&=(x+2y-2)^2+y^2+4y+3\\
&=(x+2y-2)^2+(y+2)^2-2^2+3\\
&=(x+2y-2)^2+(y+2)^2-1
\end{align*}

$x,y$は実数であるから，$(x+2y-2)^2\geqq 0,(y+2)^2\geqq 0$

したがって，$Q$は$x+2y-2=0,y+2=0$のとき最小となる．

$x+2y-2=0,y+2=0$を解くと，$x=6,y=-2$

よって，$x=6,y=-2$のとき，最小値$-1$

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% 例題I3.3.32の側注：条件なし2変数関数（One More）★★★
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% 側注1
平方完成する．

% 側注2
さらに平方完成する．

% 側注3
$a$が実数のとき，$a^2\geqq 0$であり，等号が成り立つのは，$a=0$のときであることを用いる．

% 側注4
$(\text{実数})^2\geqq 0$

% 側注5
$x,y$についての連立方程式を解く．

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% 問題I3.3.32の側注：条件なし2変数関数（One More）★★★
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% 側注1
平方完成する．

% 側注2
さらに平方完成する．

% 側注3
$a$が実数のとき，$a^2\geqq 0$であり，等号が成り立つのは，$a=0$のときであることを用いる．

% 側注4
$(\text{実数})^2\geqq 0$

% 側注5
$x,y$についての連立方程式を解く．

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% 例題I3.3.32の考え方・補足など：条件なし2変数関数（One More）★★★
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% 考え方
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(1)は，次のような手順を考えればよい．

(i) $x,y$は互いに関係なく値を取る変数であるので，先に$x$の2次式と見て$P$を平方完成し，標準形$a(x-p)^2+q$の形にする．

(ii) $y$の2次式である$q$をさらに平方完成し，標準形$b(y-r)^2+s$の形にする．

(iii) $P=a(x-p)^2+b(y-r)^2+s(a>0,b>0,s\text{は定数})$の形から，$P$は$x-p=y-r=0$のとき，最小値$s$をとる．

(2) は，$xy$の項があるが，同様の手順で$Q=a\{x-(by+c)\}^2+d(y-r)^2+s$の形に変形すればよい．