Chapter 12.3, Problem 29STP

a.

To determine

Find the volume of smallest canister.

Answer to Problem 29STP

Volume of smallest canister is $77.7i{n}^3$

Explanation of Solution

Given:

Height of smallest canister $=11inches$

Radius of smallest canister $=1.5inches$

Concept Used:

Volume of cylinder is calculated by,

  $V=\pi {\left(r\right)}^2\left(h\right)$

Where,

  $V=$ volume of cylinder.

  $r=$ radius of cylinder.

  $h=$ height of cylinder.

Calculation:

Volume of cylinder is calculated by,

  $V=\pi {\left(r\right)}^2\left(h\right)$

After putting the given values in the above equation,

  $V=\pi {\left(r\right)}^{2}h$

  $V=\pi {\left(1.5\right)}^2\left(11\right)$

  $\begin{array}{l}V=\pi \times 2.25\times 11\\ V=77.7i{n}^3\end{array}$

Conclusion:

Thus, volume of smallest canister is $77.7i{n}^3$

b.

To determine

Find the volume of medium canister.

Answer to Problem 29STP

Volume of medium canister when radius is triple of smallest canister is $683.98i{n}^3$ .

Explanation of Solution

Given:

Condition: medium canister has a triple radius that of smallest canister.

Height of medium canister $=11inches$

Radius of medium canister,

  $\begin{array}{l}=1.5\times 3\\ =4.5in\end{array}$

Concept Used:

Volume of cylinder is calculated by,

  $V=\pi {\left(r\right)}^2\left(h\right)$

Where,

  $V=$ volume of cylinder.

  $r=$ radius of cylinder.

  $h=$ height of cylinder.

Calculation:

Volume of cylinder is calculated by,

  $V=\pi {\left(r\right)}^2\left(h\right)$

After putting the given values in the above equation,

  $\begin{array}{l}V=\pi {\left(r\right)}^{2}h\\ V=\pi {\left(4.5\right)}^2\left(11\right)\\ V=\pi \times 2.25\times 11\\ V=683.98i{n}^3\end{array}$

Conclusion:

Thus, volume of medium canister when radius is triple of smallest canister is $683.98i{n}^3$ .