The current regulation ping pong ball has radius $r$ cm and requires $w$ grams of plastic. Assuming that the thickness of the balls does not change, and that the thickness is very small in comparison to the radius. The linear approximation of surface area of a ping pong ball will be $w(1+ \frac 2r Δr)$ , where $w=4\pi r^2$ is surface area.
the increase in $w$ is $≤0.06$ (6%) find largest change in radius that is allowable that will leave the price unchanged $\frac {Δr}{r}$
could you provide some hints, please?
Best Answer
Answer:
$w = 4\pi r^2$
$\delta w = 8\pi r\delta r$
$\delta w = \frac{2(\delta r) (4\pi r^2)}{r} = \frac{2(\delta r)w}{r}$
$\boxed {\delta w = \frac{2(\delta r) w}{r}}$
Now it is hits a maximum of 0.06, the price still remains the same. Now substitute the value of $\delta w$ with 0.06 and calculate $\delta r$.
$\delta r = \frac{0.06 {r}}{2\times 4\pi {r^2}}$
$\boxed{max(\delta r)= \frac{0.06}{8\pi r}}$