2006-I-15

$\begin{array}{rcl}
350 & = & k_1(50) + \dfrac{k_2 (50)^2}{500} \\
350 & = & 50k_1 +5k_2 \\
10k_1 + k_2 & = & 70~\ldots \unicode{x2460}
\end{array}$

已知 $A=20$、$n=400$ 及 $C=100$，可得

$\begin{array}{rcl}
100 & = & k_1(20) + \dfrac{k_2 (20)^2}{400} \\
100 & = & 20k_1 + k_2 \\
20k_1 + k_2 & = & 100 ~\ldots \unicode{x2461}
\end{array}$

$\unicode{x2461} – \unicode{x2460}$，可得

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

把 $k_1=3$ 代入 $\unicode{x2460}$，可得

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

所以，$C=3A+\dfrac{40A^2}{n}$。

1. 利用 (a) 的結果，可得

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

2. 由於 $P:n=5:32$，則可得

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

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

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

3. 把 $n=500$ 代入 (b)(i) 的結果，可得

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

   最大利潤

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

   所以，售出一件紀念品不能獲利 $\$100$。

4. 把 $n=400$ 代入 (b)(i) 的結果，可得

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

   所以，最大利潤為 $\$62.5$。