Q. 85 - Precalculus Enhanced with Graphing Utilities

Question

Constructing a Box: A rectangular piece of cardboard, whose area is 216 square centimeters, is made into an open box by cutting a 2-centimeter square from each corner and turning up the sides. See the figure. If the box is to have a volume of 224 cubic centimeters, what size cardboard should you start with?

Step-by-Step Solution

Answer

The required size of the cardboard is $12cm\times 18cm$.

Step 1. Form the system of equations.

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 

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

Step 2. Solve the system of equations.

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, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$,

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

which means

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

Step 3. Finding the dimensions of cardboard.

As $x=12 or x=18$ then

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

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