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NemiM [27]
3 years ago
15

A sporting goods manufacturer designs a golf ball having a volume of 2.48 cubic inches.

Mathematics
1 answer:
myrzilka [38]3 years ago
7 0
Given that the <span>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
V= \frac{4}{3} \pi r^3

The radius of the golf ball is given by
\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

For V = 2.45
</span><span>\frac{4}{3} \pi r^3=2.45 \\  \\ \Rightarrow r^3= &#10;\frac{3\times2.45}{4\pi} =0.585 \\  \\ \Rightarrow r= \sqrt[3]{0.585} &#10;=0.8363 cm

For V = 2.51
</span><span><span>\frac{4}{3} \pi r^3=2.51 \\  \\ \Rightarrow r^3= &#10;\frac{3\times2.51}{4\pi} =0.599 \\  \\ \Rightarrow r= \sqrt[3]{0.599} &#10;=0.8431 cm

Thus, </span>If the </span>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.

\lim_{r \to 0.84} V(r) =2.48
because
|V(r)-2.48|\ \textless \ 0.03
whenever
0\ \textless \ |r-0.84|\ \textless \ 0.0034
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