2006-I-15

$\begin{array}{rcl}350& =& k_1(50)+{\displaystyle \frac{k_2(50{)}^2}{500}}\\ 350& =& 50k_1+5k_2\\ 10k_1+k_2& =& 70\dots \text{①}\end{array}$

已知 $A=20$、$n=400$ 及 $C=100$，可得

$\begin{array}{rcl}100& =& k_1(20)+{\displaystyle \frac{k_2(20{)}^2}{400}}\\ 100& =& 20k_1+k_2\\ 20k_1+k_2& =& 100\dots \text{②}\end{array}$

$\text{②}–\text{①}$，可得

$\begin{array}{rcl}10k_1& =& 30\\ k_1& =& 3\end{array}$

把 $k_1=3$ 代入 $\text{①}$，可得

$\begin{array}{rcl}10(3)+k_2& =& 70\\ k_2& =& 40\end{array}$

所以，$C=3A+{\displaystyle \frac{40A^2}{n}}$。

1. 利用 (a) 的結果，可得

   $\begin{array}{rcl}P& =& 8A–C\\ & =& 8A–(3A+{\displaystyle \frac{40A^2}{n}})\\ & =& 5A–{\displaystyle \frac{40A^2}{n}}\end{array}$

2. 由於 $P:n=5:32$，則可得

   $\begin{array}{rcl}{\displaystyle \frac{P}{n}}& =& {\displaystyle \frac{5}{32}}\\ P& =& {\displaystyle \frac{5n}{32}}\end{array}$

   把 $P={\displaystyle \frac{5n}{32}}$ 代入 (b)(i) 的結果，可得

   $\begin{array}{rcl}5A–{\displaystyle \frac{40A^2}{n}}& =& {\displaystyle \frac{5n}{32}}\\ 160An–1280A^2& =& 5n^2\\ 256A^2–32An+n^2& =& 0\\ (16A-n{)}^2& =& 0\\ 16A-n& =& 0\\ 16A& =& n\\ {\displaystyle \frac{A}{n}}& =& {\displaystyle \frac{1}{16}}\\ A:n& =& 1:16\end{array}$

3. 把 $n=500$ 代入 (b)(i) 的結果，可得

   $\begin{array}{rcl}P& =& 5A–{\displaystyle \frac{40A^2}{500}}\\ & =& {\displaystyle \frac{-2}{25}}{A}^2+5A\end{array}$

   最大利潤

   $\begin{array}{cl}=& {\displaystyle \frac{4(\frac{-2}{25})(0)–5^2}{4(\frac{-2}{25})}}\\ =& {\displaystyle \frac{625}{8}}\\ <& 100\end{array}$

   所以，售出一件紀念品不能獲利 $\mathrm{\$}100$。

4. 把 $n=400$ 代入 (b)(i) 的結果，可得

   $\begin{array}{rcl}P& =& 5A–{\displaystyle \frac{40A^2}{400}}\\ & =& {\displaystyle \frac{-1}{10}}{A}^2+5A\\ & =& {\displaystyle \frac{-1}{10}}(A^2–50A)\\ & =& {\displaystyle \frac{-1}{10}}(A^2–50A+(25{)}^2–(25{)}^2)\\ & =& {\displaystyle \frac{-1}{10}}(A–25{)}^2+{\displaystyle \frac{125}{2}}\end{array}$

   所以，最大利潤為 $\mathrm{\$}62.5$。