2006-I-15

Ans: (a) $C=3A+\dfrac{40A^2}{n}$ (b) (i) $P=5A-\dfrac{40A^2}{n}$ (ii) $A:n=1:16$

(iii) Impoosible (iv) $\$62.5$

1. Let $C=k_1 A + \dfrac{k_2 A^2}{n}$, where $k_1$ and $k_2$ are non-zero constants. For $A=50$, $n=500$ and $C=350$, we have

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

   For $A=20$, $n=400$ and $C=100$, we have

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

   $\unicode{x2461} – \unicode{x2460}$, we have

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

   Sub. $k_1=3$ into $\unicode{x2460}$, we have

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

   Therefore, $C=3A+\dfrac{40A^2}{n}$.

2. 1. By the result of (a), we have

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

   2. Since $P:n=5:32$, then we have

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

      Sub. $P=\dfrac{5n}{32}$ into the result of (b)(i), we have

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

   3. Sub. $n=500$ into the result of (b)(i), we have

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

      The greatest profit

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

      Therefore, it is impossible to make a profit of $\$100$.

   4. Sub. $n=400$ into the result of (b)(i), we have

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

      Therefore, the greatest profit is $\$62.5$.