Given that the sporting goods manufacturer designs a golf ball having a volume of 2.48 cubic inches.
The golf ball is of spherical shape and the volume of a sphere is given by
[tex]V= \frac{4}{3} \pi r^3[/tex]
The radius of the golf ball is given by
[tex]\frac{4}{3} \pi r^3=2.48 \\ \\ \Rightarrow r^3= \frac{3\times2.48}{4\pi} =0.592 \\ \\ \Rightarrow r= \sqrt[3]{0.592} =0.84 cm[/tex]
For V = 2.45
[tex]\frac{4}{3} \pi r^3=2.45 \\ \\ \Rightarrow r^3=
\frac{3\times2.45}{4\pi} =0.585 \\ \\ \Rightarrow r= \sqrt[3]{0.585}
=0.8363 cm[/tex]
For V = 2.51
[tex]\frac{4}{3} \pi r^3=2.51 \\ \\ \Rightarrow r^3=
\frac{3\times2.51}{4\pi} =0.599 \\ \\ \Rightarrow r= \sqrt[3]{0.599}
=0.8431 cm[/tex]
Thus, If the ball's volume can vary between 2.45 cubic inches and 2.51 cubic inches, then the radius can vary between 0.8363 and 0.8431.
[tex] \lim_{r \to 0.84} V(r) =2.48[/tex]
because
[tex]|V(r)-2.48|\ \textless \ 0.03[/tex]
whenever
[tex]0\ \textless \ |r-0.84|\ \textless \ 0.0034[/tex]
