Chapter MPS, Problem 6.13MPS

(a)

To determine

To Find: The equation of the volume of the can.

Answer to Problem 6.13MPS

  $r^3+6r^2-160=0$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Volume of can }=160\pi i{n}^3\\ \text{Height of the can}=6+radius\text{ of can}\\ V=\pi r^{2}h\end{array}$

Formula Used:

  $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

  $\begin{array}{l}\text{Volume of can }=160\pi i{n}^3\\ \text{Height of the can}=6+radius\text{ of can}\\ V=\pi r^{2}h\\ Let,\\ \text{Radius of the can be }r\\ \text{Height of the can be}=r+6\\ Volume=\pi \left(r+6\right)r^2=160\pi \\ \text{⇒}\left(r+6\right)r^2=160\\ \text{⇒}{r}^3+6r^2-160=0\end{array}$

Conclusion:

Thus the expression for the volume of can is $r^3+6r^2-160=0$ .

(b)

To determine

To find: the possible value of radius of the can.

Answer to Problem 6.13MPS

  $4,-5-\sqrt{65}or-5+\sqrt{65}$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Volume of can }=160\pi i{n}^3\\ \text{Height of the can}=6+radius\text{ of can}\\ V=\pi r^{2}h\end{array}$

Formula Used:

  $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

  $\begin{array}{l}\text{Volume of can }=160\pi i{n}^3\\ \text{Height of the can}=6+radius\text{ of can}\\ V=\pi r^{2}h\\ Let,\\ \text{Radius of the can be }r\\ \text{Height of the can be}=r+6\\ Volume=\pi \left(r+6\right)r^2=160\pi \\ \text{⇒}\left(r+6\right)r^2=160\\ \text{⇒}{r}^3+6r^2-160=0\\ \text{factor of the 160 is 2\&5}\\ \text{putting }r=4\text{ above will become zero}\\ \text{then}\\ \text{⇒}\frac{r^3+6r^2-160}{r-4}=r^2+10r-40\\ \text{⇒}{r}^2+10r-40=0\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ a=1\\ b=10\\ c=-40\\ \text{⇒}r=\frac{-10\pm \sqrt{100+160}}{2\left(1\right)}\\ \text{⇒}r=\frac{-10\pm 2\sqrt{65}}{2}\\ \text{⇒}r=-5-\sqrt{65}or-5+\sqrt{65}\\ radius=4\\ height=4+6=10\end{array}$

Conclusion:

Thus the possible values of radius are $4,-5-\sqrt{65}or-5+\sqrt{65}$ .

(c)

To determine

To Find: The dimensions of the can.

Answer to Problem 6.13MPS

  $\begin{array}{l}radius=4in\\ height=4+6=10in\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Volume of can }=160\pi i{n}^3\\ \text{Height of the can}=6+radius\text{ of can}\\ V=\pi r^{2}h\end{array}$

Formula Used:

  $\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

  $\begin{array}{l}\text{Volume of can }=160\pi i{n}^3\\ \text{Height of the can}=6+radius\text{ of can}\\ V=\pi r^{2}h\\ Let,\\ \text{Radius of the can be }r\\ \text{Height of the can be}=r+6\\ Volume=\pi \left(r+6\right)r^2=160\pi \\ \text{⇒}\left(r+6\right)r^2=160\\ \text{⇒}{r}^3+6r^2-160=0\\ \text{factor of the 160 is 2\&5}\\ \text{putting }r=4\text{ above will become zero}\\ \text{then}\\ \text{⇒}\frac{r^3+6r^2-160}{r-4}=r^2+10r-40\\ \text{⇒}{r}^2+10r-40=0\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ a=1\\ b=10\\ c=-40\\ \text{⇒}r=\frac{-10\pm \sqrt{100+160}}{2\left(1\right)}\\ \text{⇒}r=\frac{-10\pm 2\sqrt{65}}{2}\\ \text{⇒}r=-5-\sqrt{65}or-5+\sqrt{65}\\ radius=4\\ height=4+6=10\end{array}$

Conclusion:

Thus the dimensions of the can is

  $\begin{array}{l}radius=4in\\ height=4+6=10in\end{array}$