Answer
See below:
Work Step by Step
Let the side of the square at each corner be x.
Then the height of the box made from cardboard would be x and the sides (length and width) will be $30-x-x$ or $30-2x$ as shown in the diagram.
Write the expression for the volume of the box.
$\text{Volume of box}=\text{ }\left( \text{length of box} \right)\left( \text{breadth of box} \right)\left( \text{height of box} \right)$
Substitute $V\left( x \right)$ for volume of box, $30-2x$ for length of box, $30-2x$ for breadth of box and x for height of the box.
$\begin{align}
& V\left( x \right)=\left( 30-2x \right)\left( 30-2x \right)x \\
& ={{\left( 30-2x \right)}^{2}}x \\
& =\left( 900+4{{x}^{2}}-120x \right)x \\
& =4{{x}^{3}}-120{{x}^{2}}+900x
\end{align}$
Rearrange the above expression.
$V\left( x \right)=4{{x}^{3}}-120{{x}^{2}}+900x$
Hence, the expression for the volume of the box V made by cutting the squares from each corner of the square cardboard having sides equal to $30\text{ inches}$ is $V\left( x \right)=4{{x}^{3}}-120{{x}^{2}}+900x$.
(b)
From the part (a) expression for the volume of the box V made by cutting the squares from each corner of the square cardboard having sides equal to $30\text{ inches}$ is $V\left( x \right)=4{{x}^{3}}-120{{x}^{2}}+900x$.
Consider the expression for the volume of the box.
$V\left( x \right)=4{{x}^{3}}-120{{x}^{2}}+900x$
Substitute $3$ for x.
$\begin{align}
& V\left( 3 \right)=4{{\left( 3 \right)}^{3}}-120{{\left( 3 \right)}^{2}}+900\left( 3 \right) \\
& =1738\text{ inc}{{\text{h}}^{2}}
\end{align}$
Consider the expression for the volume of the box.
$V\left( x \right)=4{{x}^{3}}-120{{x}^{2}}+900x$
Substitute $4$ for x.
$\begin{align}
& V\left( 3 \right)=4{{\left( 4 \right)}^{3}}-120{{\left( 4 \right)}^{2}}+900\left( 4 \right) \\
& =1738\text{ inc}{{\text{h}}^{2}}
\end{align}$
Consider the expression for the volume of the box.
$V\left( x \right)=4{{x}^{3}}-120{{x}^{2}}+900x$
Substitute $5$ for x.
$\begin{align}
& V\left( 3 \right)=4{{\left( 5 \right)}^{3}}-120{{\left( 5 \right)}^{2}}+900\left( 5 \right) \\
& =2000\text{ inc}{{\text{h}}^{2}}
\end{align}$
Consider the expression for the volume of the box.
$V\left( x \right)=4{{x}^{3}}-120{{x}^{2}}+900x$
Substitute $6$ for x.
$\begin{align}
& V\left( 6 \right)=4{{\left( 6 \right)}^{3}}-120{{\left( 6 \right)}^{2}}+900\left( 6 \right) \\
& =1944\text{ inc}{{\text{h}}^{2}}
\end{align}$
Consider the expression for the volume of the box.
$V\left( x \right)=4{{x}^{3}}-120{{x}^{2}}+900x$
Substitute $7$ for x.
$\begin{align}
& V\left( 2 \right)=576\left( 7 \right)+4{{\left( 7 \right)}^{3}}-96{{\left( 7 \right)}^{2}} \\
& =1792\text{ inc}{{\text{h}}^{2}}
\end{align}$
Hence, the value of $V\left( 3 \right)$ is $1728\text{ inc}{{\text{h}}^{2}}$ , value of $V\left( 4 \right)$ is $1936\text{ inc}{{\text{h}}^{2}}$ , value of $V\left( 5 \right)$ is $2000\text{ inc}{{\text{h}}^{2}}$ , value of $V\left( 6 \right)$ is $1944\text{ inc}{{\text{h}}^{2}}$ , value of $V\left( 7 \right)$ is $1792\text{ inc}{{\text{h}}^{2}}$ , and the value of the volume of the box increases for each value of x from $3$ to $5$ and then the value start decreasing.
(c)
From part (a), the expression for the volume of the box V made by cutting the squares from each corner of the square cardboard having sides equal to $30\text{ inches}$ is $V\left( x \right)=4{{x}^{3}}-120{{x}^{2}}+900x$.
Consider the expression for the volume of the box.
$V\left( x \right)=4{{x}^{3}}-120{{x}^{2}}+900x$
The length of sides to be cut from each square is x and the length of each side of square board is $30\text{ inch}$ so from each corner the length to be cut must always be less than the half of the length of the square board.
Hence, the domain for the function is:
$0 \lt x \lt 15$
