# Circular metal disk…“ideal” radius (Epsilon and delta)

At Orange County Choppers, Paul Teutul Junior needs to make a circular metal disk with area 825 in^2.

The radius of such a disk is _ inches.

To keep Paul Senior from blowing a gasket, Paul Junior must deviate from the ideal area of the disk, which is 825 in^2, by less than +/-4 in^2. How close to the ideal radius must the Flowjet (the machine that cuts the disk) be to maintain tranquility at OCC?

In terms of the (epsilon), (delta) definition of lim->{x to a} f(x) = L, let x be the actual radius of the disk and f(x) the actual area of the disk. What is the formula for the function f(x)? What is the number a? What is the number L? What value of (epsilon) is given? What is the corresponding value of (delta)?

This is a HUGE problem...I got the first part, which was easy. The radius is sqrt[825/pi]. However, I do NOT know how to get the answer to the second part, because I don't know how to set up the problem. I tried subtracting and adding 4 to the ideal area, but all I got for the radius's was sqrt[821/pi] and sqrt[829/pi], which, when you plug those in for the Area equation, both have equal distances from the ideal radius. So.....??? I'm confused.

Your thought, to check $\sqrt{\frac{829}{\pi}}$ and $\sqrt{\frac{821}{\pi}}$ compared with $\sqrt{\frac{825}{\pi}}$ is correct. But they are not exactly the same distance from $\sqrt{\frac{825}{\pi}}.$ To five places, $\sqrt{\frac{829}{\pi}}-\sqrt{\frac{825}{\pi}}=0.39324$, while $\sqrt{\frac{825}{\pi}}-\sqrt{\frac{821}{\pi}}=0.39333.$ In this case $\epsilon$ is the $4 \text{ in}^2$ of allowable error. So to maintain tranquility you take $\delta$ as the minimum of these two deltas. The point of the exercise is that the allowable delta radius is a slowly varying function of the allowable delta area. There may be a future problem where it goes the other way-where the tolerance on one side is much greater than the tolerance on the other.