【数学I】例題3.2.10：2次関数の最大・最小の文章題（One More）★★

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% 例題I3.2.10：2次関数の最大・最小の文章題（One More）★★
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周囲の長さが$12$である長方形において，次の問いに答えよ．

(1) この長方形の面積の最大値を求めよ．

(2) この長方形の対角線の長さの最小値を求めよ．

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% 例題I3.2.10の解答（One More）★★
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(1) 長方形の縦の長さを$x$とすると，

横の長さは$\frac{12-2x}{2}=6-x$である．

また，$x>0,6-x>0$であるから，$0<x<6$

この長方形の面積を$y$とすると，

\[
y=x(6-x)=-x^2+6x=-(x-3)^2+9
\]

$0<x<6$より，$y$は$x=3$で最大値9をとる．

よって，長方形の面積の最大値は，$9$

(2) 長方形の対角線の長さを$l$とすると，

\[
l^2=x^2+(6-x)^2=2x^2-12x+36=2(x-3)^2+18
\]

$0<x<6$において，$l^2$は$x=3$で最小値$18$をとる．

ここで，$l>0$であるから，$l^2$が最小となるとき，$l$も最小となる．

よって，対角線の最小値は，$\sqrt{18}=3\sqrt{2}$

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% 問題I3.2.10：2次関数の最大・最小の文章題（One More）★★
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直角を挟む2辺の長さの和が10である直角三角形において，斜辺の長さが最小となる直角三角形を求め，その斜辺の長さを求めよ．

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直角を挟む2辺のうち一方の辺の長さを$x$とすると，他方の辺の長さは$10-x$である．

また，$x>0$，$10-x>0$であるから，$0<x<10$

斜辺の長さを$l$とすると，

\[
l^2=x^2+(10-x)^2=2x^2-20x+100=2(x^2-10x)+100=2(x-5)^2+50
\]

$0<x<10$において，$l^2$は$x=5$で最小値50をとる．

このとき，他方の辺の長さは$10-5=5$

ここで，$l>0$であるから，$l^2$が最小となるとき，$l$も最小となる．

よって，求める直角三角形は，直角を挟む2辺の長さがともに5の直角二等辺三角形で，斜辺の長さは，$\sqrt{50}=5\sqrt{2}$

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% 例題I3.2.10の側注：2次関数の最大・最小の文章題（One More）★★
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% 側注1
辺の長さが正であることから，$x$の値の範囲を求める．

% 側注2
三平方の定理を用いる．

% 側注3
$a>0,b>0$のとき，

\[
a<b\Longleftrightarrow a^2<b^2
\]
を用いている．詳しくは，数学IIで学習する．

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% 例題I3.2.10の考え方・補足など：2次関数の最大・最小の文章題（One More）★★
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長方形の縦の長さと横の長さを$x$を用いて表すと，長方形の面積は$x$の関数となる．縦の長さを$x$とすると，横の長さは$\frac{12-2x}{2}=6-x$となる．なお，最大値・最小値を求めるときは，$x$の値の範囲に注意すること．

(2) 対角線の長さ$l$は，根号を含む$l=\sqrt{f(x)}$の形となる．このようなときは，$l^2=f(x)$の最小値を求めて，$l$の最小値を考える．

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文字の値の範囲に注意すること．