2012PP-I-11

When $x=20$, $C=42$, we have

$42=k_1+400k_2 ~ \ldots \unicode{x2460} $

When $x=120$, $C=112$, we have

$112=k_1+14400k_2 ~\ldots \unicode{x2461}$

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

$\begin{array}{rcl}
70 & = & 14000k_2 \\
k_2 & = & \dfrac{1}{200}~\ldots \unicode{x2462}
\end{array}$

Sub. $\unicode{x2462}$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
42 & = & k_1+400(\dfrac{1}{200}) \\
k_1 & = & 40
\end{array}$

Therefore $C=40+\dfrac{x^2}{200}$.

For $x=50$, we have

$\begin{array}{rcl}
C & = & 40 + \dfrac{50^2}{200} \\
& = & 52.5
\end{array}$

Therefore, the required cost is $\$52.5$.

$\begin{array}{rcl}
58 & = & 40 +\dfrac{x^2}{200} \\
x^2 & = & 3600 \\
x & = & 60 \mbox{ or }-60\mbox{ (rejected)}
\end{array}$

Therefore, the required length is $60\mbox{ cm}$.