Ans: (a) $C=4x+5x^2$ (b) $\dfrac{36}{5}\text{ m}$
- Let $C=k_1x + k_2x^2$, where $k_1$ and $k_2$ are non-zero constants. For $x=4$ and $C=96$, we have
$\begin{array}{rcl}
96 & = & k_1(4) +k_2(4)^2 \\
24 & = & k_1 + 4k_2~\ldots\unicode{x2460}
\end{array}$For $x=5$ and $C=145$, we have
$\begin{array}{rcl}
145 & = & k_1(5) + k_2 (5)^2 \\
29 & = & k_1 + 5k_2 ~\ldots\unicode{x2461}
\end{array}$$\unicode{x2461}-\unicode{x2460}$, we have
$\begin{array}{rcl}
5 & = & k_2 \\
k_2 & = & 5
\end{array}$Sub. $k_2=5$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
24 & = & k_1 + 4(5) \\
k_1 & = & 4
\end{array}$Therefore, $C=4x+5x^2$.
- For $C=288$, we have
$\begin{array}{rcl}
288 & = & 4x+5x^2 \\
5x^2 + 4x -288 & = & 0 \\
(5x-36)(x+8) & = & 0 \\
\therefore x=\dfrac{36}{8} & \text{or} & x=-8\text{(rejected)}
\end{array}$Therefore, the perimeter is $\dfrac{36}{8}\text{ m}$.
