After how many years will the maximum payout be less than the premium, losing money by continuing with the policy in the scenario below?A machine...

The annual premium paid is 10% of the new value = P/10 ,
where P is the new value of the machine.

The  value of the
machine due to depreciation  is 1/2 of new value for every two years. Thus the machine
value at any year becomes sqrt(1/2) of its pevious year value.  The new value of the
machine is P . So after n years the value of the machine becomes  {sqrt(1/2)} ^n * P=
P/(2^(n/2)).

So the maximum pay out by the insurance
company  at any year = machine value at that time (or year) =
P/(2^(n/2)).

So it is required to determine when the
maximum pay out by the insurance company becomes less than the annual premum
.

Therefore to determine n such that P/2^(n/2)  <
P/10.

Divide by P and cross
multiply:

10 <
2^n/2.

log10 < (n/2)
log2.

2 /log2 > n.

Or n
> 2/log2 .

Clearly for n = 6.64 ,  1/2(6.64/2) =
0.100226.. > 1/10.

For n = 6.65 , 1/2^(6-65/2) =
0.9979 < 1/10.

Therefore  between the year 6 and 7 
(or between 6.64 years and 6.65 years, or in the 8th month  after 6 years) the  maximum
 insurance becomes less than the annual premium. So making the annual payment of premium
is a clear loss from the 7th year onwards.