Question

You are making a box from a 5-inch square piece of cardboard. The box will be formed by cutting out the comets as shown in the diagram and folding up the sides. You want the volume of the box to be 2 cubic inches. Find the rational solution(s) of the equation. Then use polynomial long division to find the other solution(s). What are the possible side lengths of the box ?

Answer 1

From the square piece of cardboard, the length of the corners that were cut out would be:

`(5-x)/(2)`

On folding the cardboard to form a box, the dimensions of the box now become:

`\begin{gathered} Length(L)=x \\ Width(W)=x \\ Height(H)=(5-x)/(2) \end{gathered}`

If the volume of the box is to be 2 cubic inches, therefore,

`\begin{gathered} Volume\text{ of box = }Length(L)* Width(W)* Height(H) \\ =x* x*(5-x)/(2) \\ =x^2((5-x)/(2)) \\ \text{Equating the volumes,} \\ (x^2(5-x))/(2)=2 \\ x^2(5-x)=4 \\ 5x^2-x^3=4 \\ R\text{earranging it,} \\ x^3-5x^2+4=0 \end{gathered}`

By trial and error method,

`\begin{gathered} x=1\text{ is a solution} \\ \text{Hence, (}x-1)\text{ is one of the factors} \end{gathered}`

Using the polynomial long division method to find the other solution

The quotient obtained from the long division is

`x^2-4x-4=0`

Using the quadratic formula to solve for x

`\begin{gathered} x=(-b\pm√(b^2-4ac))/(2a) \\ \text{where, }a=1,\text{ }b=-4,\text{ }c=-4 \\ x=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(1)(-4)}}{2(1)} \\ x=\frac{4\pm\sqrt[]{16^{}+16}}{2(1)} \\ x=\frac{4\pm\sqrt[]{32}}{2} \\ x=\frac{4\pm4\sqrt[]{2}}{2} \\ x=2+2\sqrt[]{2}\text{ = }=2(1+\sqrt[]{2})\text{ =}4.828 \\ x=2-2\sqrt[]{2}\text{ = }=2(1-\sqrt[]{2})=-0.828 \end{gathered}`

Hence, all the rational solutions of the equation are:

`\begin{gathered} x=1 \\ x=4.828 \\ x=-0.828 \\ \text{Note that a side length cannot be negative, so, }x\\e-0.828 \end{gathered}`

The possible side lengths of the box would be:

`\begin{gathered} If\text{ }x=1 \\ Length(L)=1\text{ in} \\ Width(W)=1\text{ in} \\ Height(H)=(5-1)/(2)=(4)/(2)=2\text{ in} \end{gathered}`
`\begin{gathered} If,x=4.828 \\ Length(L)=4.828\text{ in} \\ Width(W)=4.828\text{ in} \\ Height(H)=(5-4.828)/(2)=0.086\text{ in} \end{gathered}`