Anew 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.

The answer is 15×7=105therefore, 105×5=525 miles