A new restaurant estimates that if it has seats for 25 to 50 people, it will make $10 per seat in daily profit. If the restaurant could seat more than 50 people, the daily profit will be decreased by $0.05 per person beyond 50. How many should the restaurant seat in order to maximize its daily profits?

Let's

x=people beyond 50

x+50=maximun number of seats

10-0.05 (x) = profit per seat.

We have the following function:

P (x) = profits

P (x) = (x+50) (10-0.05x)

P (x) = 10x-0.05x²+500-2.5x

P (x) = - 0.05x²+7.5x+500

1) we have to compute the first derivative:

P' (x) = - 0.1x+7.5

2) we find the values of "x", when P' (x) = 0:

P' (x) = 0

-0.1x+7.5=0

x=-7.5 / (-0.1)

x=75

3) we have to compute the second derivative to check out if 75 is a maximum value or minimum value.

P'' (x) = - 0.1<0; then in x=75, we have a máximum.

4) we compute the maximum number of seats

x+50=75+50=125

The number of seats would have to be 125.

And the profit would be:$781.25

P (x) = (x+50) (10-0.05x) = 125 (6.25) = $781.25

50. it'll decrease by $10 per seat if it's any lower, and it'll decrease by $0.05 per seat if it's any higher. 50 people seated will give them the most possible daily profits.