Chapter 6.8, Problem 18PPS

(a)

To determine

To find:the function $V\left(x\right)$ for the volume of the box.

Answer to Problem 18PPS

  $V\left(x\right)={\left(28-2x\right)}^2.x$ .

Explanation of Solution

Given:

The width as $28\text{ cm}$ and by making the square sheet to box it cut the $4$ small square form the $4$ corners of the square sheet.

  

Concept used:

The volume of the square is:

  $a^3$ .

Calculation:

According to the given diagram which is in the form of square box having its total length and width as $28\text{ cm}$ and by making the square sheet to box it cut the $4$ small square form the $4$ corners of the square sheet.

Let $h=x$

Then the length and width will be:

  $\begin{array}{l}l=28-2x.\\ w=28-2x.\end{array}$

Length and width will be equal both are $28$ minus the distance which is cut $x$ after being folded.

Height will be the value of $x$ .

The volume of the box will be:

Hence, $V\left(x\right)={\left(28-2x\right)}^2.x$ .

(b)

To determine

To find:the length of the cut square $x$ when the volume of the box is equal to $1152{\text{cm}}^3$ .

Answer to Problem 18PPS

  $x\in \left\{2,8\right\}$ .

Explanation of Solution

Given:

The volume of the box equals to $1152$ cubic centimetres.

Concept used:

In synthetic substitution method:

If $f\left(x\right)=a{x}^2+bx+c$ .

  $r\overline{)\begin{array}{l}\underset{\_}{\begin{array}{l}abc\\ ara{r}^2+br\end{array}}\\ aar+ba{r}^2+br+c\end{array}}$

Then the last remainder is the solution of the function.

  $f\left(r\right)=a{r}^2+br+c$ .

Calculation:

  $V\left(x\right)={\left(28-2x\right)}^2.x$ .

According to the given the volume of the box equals to $1152$ cubic centimetres.

  $\begin{array}{l}V\left(x\right)=1152.\\ {\left(28-2x\right)}^{2}x=1152.\end{array}$

The equation can be solved by multiplying the terms:

  $\begin{array}{l}{\left(28-2x\right)}^{2}x=1152.\\ 4x^3-112x^2+784x-1152=0.\\ x^3-28x^2+196x-288=0.\end{array}$

At $x=2$ ; $f\left(x\right)=0$ .

By using synthetic substitution methods to find the depreciated polynomial at $x=2$ :

  $2\overline{)\begin{array}{l}\underset{\_}{\begin{array}{l}1-28196-288\\ 2-52288\end{array}}\\ 1-261440\end{array}}$

The equation will be:

  $\begin{array}{l}\left(x-2\right)\left(x^2-26x+144\right)=0.\\ \left(x-2\right)\left(x-18\right)\left(x-8\right)=0.\\ x_1=8,x_2=18\text{ and }{x}_3=8.\end{array}$

The length of the square side is $28-2x$ .

To check which solution is fit:

  $\begin{array}{l}28-2x>0.\\ 28>2x.\\ x<14.\end{array}$

Therefore, the only two solution is fit for the solution as:

Hence, $x\in \left\{2,8\right\}$ .

(c)

To determine

To find:the volume of the box having length of the square cut $x=6$ centimetre.

Answer to Problem 18PPS

  $V\left(6\right)=1536.$

Explanation of Solution

Given:

  $x=6$ .

Concept used:

The volume of the square is:

  $a^3$ .

  $\begin{array}{l}l=28-2x.\\ w=28-2x.\end{array}$

Calculation:

  $V\left(x\right)={\left(28-2x\right)}^2.x$

The volume of the function when $x=6$ is:

  $\begin{array}{l}V\left(x\right)={\left(28-2x\right)}^2.x.\\ V\left(6\right)={\left(28-2\left(6\right)\right)}^2.6=256\times 6=1536.\end{array}$

Hence, $V\left(6\right)=1536.$