How to calculate the ideal maintaining time of a machine?

Suppose the probability of an error in the $n$th cycle since the most recent maintenance is $a+bn$, and let the number of tasks (cycles) between maintenance be $x$. I'm assuming here that the tasks can be treated as discrete.

The way I would approach the optimization is to calculate both the total maintenance cost and the expected total error cost over some fairly large number of cycles - say 1,000. What number is chosen here is arbitrary, but I find it more intuitive to think of total costs in that context than averages per cycle, although each should lead to the same result.

Total maintenance costs are:

$$frac{1000}{x}times 1000=frac{1000000}{x}$$

Expected error costs between two successive maintenance events, using the triangle number formula to sum the $b$'s, are:

$$sum_{n=1}^x 15(a+bn)=15Big(ax+bBig(frac{x(x+1)}{2}Big)Big)$$

Hence total expected error costs are:

$$frac{1000}{x}times 15Big(ax+bBig(frac{x(x+1)}{2}Big)Big)=15000Big(a+Big(bfrac{x+1}{2}Big)Big)$$

Total costs $TC$ (including both maintenance and expected error costs) are therefore:

$$TC = frac{1000000}{x} + 15000Big(a+Big(bfrac{x+1}{2}Big)Big)$$

Setting the first derivative equal to zero to find the minimum:

$$frac{dTC}{dx}=frac{-1000000}{x^2}+15000Big(frac{b}{2}Big)=0$$

$$-2000000+15000bx^2=0$$

$$-400+3bx^2=0$$

$$x=sqrt{frac{400}{3b}}$$

To confirm this is a minimum:

$$frac{d^2TC}{dx^2}=frac{2000000}{x^3}>0$$

Putting (as approximately suggested by the chart) $a=0.18, b=0.001$, this yields:

$$x=sqrt{frac{400}{0.003}}approx365$$

Note that in this case $a+bx=0.18+(0.001times365)=0.545 < 1$. The formula would need modifying should the values imply a probability of error exceeding 1 before reaching the next maintenance event.

Addendum 3/11/2020

In response to your comment re integration, if the tasks are taken to be discrete then integration does not give an exactly correct formula for the total expected error costs. The integral needed is:

$$int_0^x 15(a+bn)dn=15Big(ax+frac{bx^2}{2}Big)$$

This has $x^2$ where the calculation above has $x(x+1)$. However, this difference vanishes on differentiating TC since:

$$frac{d(bx)}{dx}=frac{d(b(x+1))}{dx}=b$$

Thus this approach leads to the same optimum value for $x$.