From the problem you are posing it is obvious we need to find the profit below 450 and also above 450.

The below 450 case.

The number of registrations is easier to compute, and then we can solve for f, the registration fee.

`x = 450 - (f-25)/0.50*10 = 450...

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From the problem you are posing it is obvious we need to find the profit below 450 and also above 450.

The below 450 case.

The number of registrations is easier to compute, and then we can solve for f, the registration fee.

`x = 450 - (f-25)/0.50*10 = 450 - 20f + 500 = 950 - 20f` , so `f = 1/20(950-x) = 47.50-x/20`

so the profit `P(x) = (47.50 - x/20)x - 118 - 15x = -x^2/20 + 32.50x - 118`

The marginal profit `(dP(x))/(dx) = -2x/20 + 32.50 = 0` when x = 325, and the profit there is `P(325) = -(325)^2/20 + 32.50(325) - 118 = $5163.25` . Registration Fee = 31.25

We need to check the end points [0, 449], P(0) = -118, and P(449) = 4395. So x = 325 is a maximum.

Now for x>=450. `P(x) = 25x - 11x - 118 = 14x - 118` . Now there are no critical points, but we can find when P(x) > $5163.25. This is at x = 337. At 450 they would make $6182.

So to maximize we should charge $25 and expect more than 450 people registered.