Answer
(a) $r = 17.84~cm$
(b) The radius must be within $\pm 0.04~cm$
(c) $x$ is the radius
$f(x) = \pi~x^2$
$a$ is the ideal radius of $17.84~cm$
$L$ is the area of $1000~cm^2$
$\epsilon = 5~cm^2$
$\delta = 0.04~cm$
Work Step by Step
(a) We can find the required radius:
$\pi~r^2 = 1000~cm^2$
$r^2 = \frac{1000~cm^2}{\pi}$
$r = \sqrt{\frac{1000~cm^2}{\pi}}$
$r = 17.84~cm$
(b) We can find the required radius if $A = 995~cm^2$:
$\pi~r^2 = 995~cm^2$
$r^2 = \frac{995~cm^2}{\pi}$
$r = \sqrt{\frac{995~cm^2}{\pi}}$
$r = 17.80~cm$
We can find the required radius if $A = 1005~cm^2$:
$\pi~r^2 = 1005~cm^2$
$r^2 = \frac{1005~cm^2}{\pi}$
$r = \sqrt{\frac{1005~cm^2}{\pi}}$
$r = 17.89~cm$
The radius must be within $\pm 0.04~cm$
(c) $\lim\limits_{x \to a}f(x) = L$
$x$ is the radius
$f(x) = \pi~x^2$
$a$ is the ideal radius of $17.84~cm$
$L$ is the area of $1000~cm^2$
$\epsilon = 5~cm^2$
$\delta = 0.04~cm$
