Checkpoint: Applications for Rational Expressions Solution

Find the efficiency ratio $\frac{S A}{V}$ that compares the surface area and volume of a sphere.

$\begin{matrix} S A = 4 π r^{2} & V = \frac{4}{3} π r^{3} \end{matrix}$

$\begin{array}{rcl} \frac{S A}{V} & = & \frac{4 π r^{2}}{\frac{4}{3} π r^{3}} = \frac{4}{\frac{4}{3} r} \\ = & 4 \div \frac{4 r}{3} = \frac{4}{1} \cdot \frac{3}{4 r} \\ \frac{S A}{V} & = & \frac{3}{r} \end{array}$

Note

Q: What are the restrictions on the radius?
A: The radius has to be greater than zero.

The radius cannot be zero because it is in the denominator. The radius must be greater than zero because the radius cannot be a negative value.