Question 1067268
Let n = number of nickels, q = number of quarters.

Then we have  n+q =16  and  .05n+.25q >= 2.20

Notice that we have two other constraints.   0<=n=<16 and 0<=q<=16

So  the solution that gives us exactly 16 coins and exactly 2.20 would be set by

n+q=16 and .05n+.25q=2.20

q=16-n by solving the 1st equation for q.

.05n + .25(16-n) = 2.20

.05n + 4 - .25n = 2.20

.2n = 1.80

n = 9  (so q = 7)

So n is at least 9.

Now, the question is what is the greatest?

We can logically say that for every nickel I add, I have to subtract a quarter (meaning the net amount of money is decreased by 20 cents).

So for instance if n = 10, then q = 6, so we have .5 + 1.50 = 2.00 (which isn't at least 2.20). 

Now we can have n=8 and we will go to 2.40, but that's not the "maximum" number of nickels.

So, the maximum number of nickels is 9.

I'll attempt to graph what is going on.

{{{graph(500,500,0,16,0,16,y=16-x,y>=8.8-.2x)}}}

This is using q = 16-n
and  q >= 8.8-.2n
Notice that the intersection of the graphs for (n,q) is (9,7)