Question

A rectangular piece of cardboard is a=9 in. longer than it is wide. Squares 2 in. on a side are to be cut from each corner and then the sides folded

up to make an open box with a volume of 72 in³. Find the length and width of the piece of cardboard.

Answer 1

Length= 16 inches

Width= 7 inches

Explanation:

Let's denote the width of the rectangular piece of cardboard as "w" inches.

According to the given information,

the length of the cardboard is 9 inches longer than its width, so the length can be represented as "w + 9" inches.

When the sides are folded up, the height of the box will be 2 inches.

After folding, the width of the base of the box will be "w - 4" inches, and the length will be "w + 9 - 4" inches, which simplifies to "w + 5" inches.

The volume of a rectangular box can be calculated as the product of its length, width, and height.

In this case, the volume is given as 72 in³:

Volume = Length*Width*Height

72 = (w + 5)*(w - 4) × 2

Simplifying the equation:

36 = (w + 5)*(w - 4)

Expanding the right side:

36 = w² - 4w + 5w - 20

Rearranging and combining like terms:

w² + w - 56 = 0

We can factorize by using middle term factorization:

w² + 8x-7x - 56 = 0

w(w+8)-7(w+8)=0

(w - 7)(w + 8) = 0

Setting each factor equal to zero:

either

w - 7 = 0

Therefore, w = 7

or

w + 8 = 0

w = -8 (Discard since width cannot be negative)

Therefore, the width of the piece of cardboard is 7 inches.

Substituting this value back into the expression for the length:

Length = w + 9

Length = 7 + 9

Length = 16 inches

So, the length of the piece of cardboard is 16 inches, and the width is 7 inches.

Answer 2

Length = 16 inches

Width = 7 inches

Explanation:

Let "x" be the width of the rectangular piece of cardboard.

If the rectangular piece of cardboard is 9 inches longer than it is wide, then let "x + 9" be the length of the rectangular piece of cardboard.

As squares with side lengths of 2 inches are cut from each corner of the rectangle, the height of the open box will be 2 inches, and the width and length of the box will be 4 inches less than width and length of the rectangular piece of cardboard.

Therefore, the dimensions of the open box are:

• Height, h = 2 inches
• Width, w = (x - 4) inches
• Length, l = (x + 9 - 4) = (x + 5) inches

Given the open box has a volume of 72 in³, substitute all the values into the formula for the volume of a cuboid and solve for x:

`\begin{aligned}\textsf{Volume of a cuboid}&=\sf width \cdot length \cdot height\\\\72&=(x-4)\cdot (x+5)\cdot 2\\36&=(x-4)(x+5)\\(x-4)(x+5)&=36\\x^2+x-20&=36\\x^2+x-56&=0\\x^2+8x-7x-56&=0\\x(x+8)-7(x+8)&=0\\(x-7)(x+8)&=0\\\\x-7&=0 \implies x=7\\x+8&=0 \implies x=-8\end{aligned}`

As length cannot be negative, the only valid value of x is 7.

To find the length and width of the piece of cardboard, substitute the found value of x into their expressions:

`\textsf{Length}=x+9=7+9=16\; \sf inches`

`\textsf{Width}=x=7\; \sf inches`

Therefore, the dimensions of the rectangular piece of cardboard are:

• Length = 16 inches
• Width = 7 inches