91Ó°ÊÓ

Let x be the length of the rectangular piece of cardboard and let y be its breadth.

The area of the rectangular piece is 216 square centimeters. Thus, the first equation is

$xy=216$

Next, The box is formed by cutting a 2 centimeter square from each corner.

So, the length and the breadth are reduced by 4 .

The new length and breadth are $(x-4),(y-4)$respectively.

The height of the box will be 2 centimeters.

The volume of the box is 224 cubic centimeters.

So, the second equation is $(x-4)(y-4)2=224$.

Thus, the system of equations is

$$\begin{gathered} xy=216 \\ 2(x-4)(y-4)=224 \end{gathered}$$

From first equation,

$$\begin{gathered} xy=216 \\ \frac{xy}{x}=\frac{216}{x} \\ y=\frac{216}{x} \end{gathered}$$

From second equation,

$$\begin{gathered} 2(x-4)(y-4)=224 \\ \frac{2(x-4)(y-4)}{2}=\frac{224}{2} \\ (x-4)(y-4)=112 \\ xy-4x-4y+16=112 \\ xy-4x-4y+16-112=0 \\ xy-4x-4y-96=0 \\ x\left(\frac{216}{x}\right)-4x-4\left(\frac{216}{x}\right)-96=0(y=\frac{216}{x}) \\ 216-4x-\frac{864}{x}-96=0 \\ -4x-\frac{864}{x}+120=0 \\ -x-\frac{216}{x}+30=0 \\ -x^2-216+30x=0 \\ {x}^2-30x+216=0 \end{gathered}$$

Using the quadratic formula, $x=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a}$,

$$\begin{gathered} x=\frac{30Â\pm \sqrt{{(-30)}^2-4\left(1\right)\left(216\right)}}{2\left(1\right)} \\ =\frac{30Â\pm \sqrt{900-864}}{2} \\ =\frac{30Â\pm 6}{2} \end{gathered}$$

which means

$$\begin{gathered} x=\frac{30+6}{2}orx=\frac{30-6}{2} \\ x=18orx=12 \end{gathered}$$

As $x=12orx=18$then

$$\begin{gathered} y=\frac{216}{x} \\ y=\frac{216}{12}ory=\frac{216}{18} \\ y=18ory=12 \end{gathered}$$

So the required size of the cardboard is$12cm×18cm$.