2013-I-11

For $\ell=1$ and $W=181$, we have

$\begin{array}{rcl}
k_1(1)+k_2(1)^2 & = & 181\\
k_1+k_2 & = & 181 \ \ldots \unicode{x2460}
\end{array}$

For $\ell=2$ and $W=402$, we have

$\begin{array}{rcl}
k_1(2)+k_2(2)^2 & = & 402 \\
k_1 + 2k_2 & = & 201 \ \ldots \unicode{x2461}
\end{array}$

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

$\begin{array}{rcl}
k_2 & = & 20 \\
\end{array}$

Sub. $k_2=20$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
k_1 + 20 & = & 181 \\
k_1 & = & 161
\end{array}$

Therefore, $W=161\ell+20\ell^2$. For $\ell=1.2$, we have

$\begin{array}{rcl}
W & = & 161(1.2) + 20(1.2)^2 \\
& = & 222
\end{array}$

Therefore the weight of a tray of perimeter $1.2$ metres is $222$ grams.

$\begin{array}{rcl}
594 & = & 161\ell + 20 \ell^2 \\
20\ell^2 + 161\ell – 594 & = & 0 \\
(5\ell+54)(4\ell – 11) & = & 0 \\
\ell = \dfrac{-54}{5}\text{ (rejected)} & \text{or}& \ell = \dfrac{11}{4}
\end{array}$

Therefore, the required perimeter is $\dfrac{11}{4}$ metres.