Chapter 5.4, Problem 53PS

Solution

To prove: The ratio of the surface area to the volume of the container is $\frac{6\left(2r+l\right)}{r\left(4r+3l\right)}$ .

Given information:

The length of the cylinder: $l$

The radius of each hemisphere: $r$

Formula used:

Volume and surface area of sphere:

  $V_S=\frac{4}{3}\pi r^3;S_s=4\pi r^2$

Volume and surface area of cylinder:

  $V_c=\pi r^{2}h;S_s=2\pi r^2+2\pi rh$

Proof:

The two objects, a sphere (two hemispheres) and a cylinder.

Write out the volume and surface area formulas for each object:

  $\begin{array}{l}{V}_S=\frac{4}{3}\pi r^3;S_s=4\pi r^2\\ V_c=\pi r^{2}h;S_s=2\pi r^2+2\pi rh\end{array}$

The surface area of the object,

  $\begin{array}{l}4\pi r^2+2\pi r^2+2\pi rl-\pi r^2-\pi r^2\\ 4\pi r^2+2\pi rl\end{array}$

The volume of the object;

  $\frac{4}{3}\text{}\pi r^3+\pi r^{2}l$

Ratio of surface area to volume for the object:

  $\frac{4\pi r^2+2\pi rl}{\frac{4}{3}\text{}\pi r^3+\pi r^{2}l}$

Multiply by $3$ ;

  $\frac{12\pi r^2+6\pi rl}{4\text{}\pi r^3+3\pi r^{2}l}$

Factor the greatest common factor from the numerator.

Factor the greatest common factor from the denominator.

  $\frac{6\left(2r+l\right)}{r\left(4r+3l\right)}$

Hence, proved