Answer
Length: 18 inches
Width: 12 inches
Work Step by Step
Let's note:
$l$=the length of the cardboard
$w$=the width of the cardboard
The dimensions of the box after removing 2 inches from each corner are:
Length: $l-2(2)=l-4$
Width: $w-2(2)=w-4$
Height: $2$
We can write the system using the area of the cardboard and the volume of the box:
$\begin{cases}
lw=216\\
(l-4)(w-4)2=224
\end{cases}$
$\begin{cases}
lw=216\\
(l-4)(w-4)=112
\end{cases}$
We will use the substitution method. Solve Equation 1 for $w$ and substitute the expression of $w$ in Equation 2 to eliminate $w$ and determine $l$:
$\begin{cases}
w=\dfrac{216}{l}\\
(l-4)\left(\dfrac{216}{l}-4\right)=112
\end{cases}$
$(l-4)\cdot \dfrac{216-4l}{l}=112$
$(l-4)(216-4l)=112l$
$4(l-4)(54-l)=112l$
$(l-4)(54-l)=28l$
$54l-l^2-216+4l=28l$
$-l^2+58l-216-28l=0$
$-l^2+30l-216=0$
$l^2-30l+216=0$
$l^2-12l-18l+216=0$
$l(l-12)-18(l-12)=0$
$(l-12)(l-18)=0$
$l-12=0\Rightarrow l_1=12$
$l-18=0\Rightarrow l_2=18$
Substitute each of the values of $l$ in Equation 1 to determine $w$:
$w=\dfrac{216}{l}$
$l_1=12\Rightarrow w=\dfrac{216}{12}\Rightarrow w_1=18$
$l_2=18\Rightarrow w=\dfrac{216}{18}\Rightarrow w_2=12$
The system's solutions are:
$(12,18),(18,12)$
As $l\geq w$, the solution is:
$l=18$ inches
$w=12$ inches
