12-8 할리데이 11판 솔루션 일반물리학

$$\ab{a}$$ $$\com(1)={L\over2}\le L-a_1,$$ $$ \begin{aligned} a_1\le{L\over2}\taag1 \end{aligned} $$ $$\therefore \max(a_1)={L\over2}$$ $$\ab{b}$$ $$\com(12)={L+a_1\over2}\le L-a_2,$$ $$ \begin{aligned} a_2\le{L-a_1\over2}\taag2 \end{aligned} $$ $$\therefore \max(a_2)={L\over2}\\(a_1=0)$$ $$\ab{c}$$ $$\com(123)={L+a_1+a_2\over2}\le L-a_3,$$ $$ \begin{aligned} a_3\le{L-a_1-a_2\over2}\taag3 \end{aligned} $$ $$\therefore \max(a_3)={L\over2}\\(a_1=0, a_2=0)$$ $$\ab{d}$$ $$\com(1234)={L+a_1+a_2+a_3\over2}\le L-a_4,$$ $$ \begin{aligned} a_4\le{L-a_1-a_2-a_3\over2}\taag4 \end{aligned} $$ $$\therefore \max(a_4)={L\over2}\\(a_1=0, a_2=0,a_3=0)$$ $$\ab{e}$$ $$h=a_1+a_2+a_3+a_4,$$ $$h\le a_1+{L-a_1\over2}+{L-a_1-a_2\over2}+{L-a_1-a_2-a_3\over2}\\ $$ $$h\le{a_1+7L\over8}$$ $$ h\le{15\over16}L\\ $$ $$\therefore \max(h)={15\over16}L\\\br{a_1={L\over2}, a_2={L\over4},a_3={L\over8},a_4={L\over16}}$$