A local club is arranging a charter flight to Hawaii. The cost of the trip is ​$586 each for 80 ​passengers, with a refund of​ $5 per passenger for each passenger in excess of 80.(a) Find the number of passengers that will maximize the revenue received from the flight.(b) Find the maximum revenue.

Answer:a) The number of passengers that will maximize the revenue received from the flight is 99.b) The maximum revenue is $48,609.Step-by-step explanation:We have to analyse two cases to build a piecewise function.If there are 80 or less passengers, we have that:The cost of the trip is $586 for each passenger. So[tex]R(n) = 586n[/tex]If there are more than 80 passengers.There is a refund of $5 per passenger for each passenger in excess of 80. So the cost for each passenger is[tex]R(n) = (586 - 5(n-80))n = -5n^{2} +400n + 586n = -5n^{2} + 986n[/tex].So we have the following piecewise function:[tex]R(n) = \left \{ {{586n}, n\leq 80 \atop {-5n^{2} + 986n}, n > 80} \right[/tex]The maxium value of a quadratic function in the format of [tex]y(n) = an^{2} + bn + c[/tex] happens at:[tex]n_{v} = -\frac{b}{2a}[/tex]The maximum value is:[tex]y(n_{v})[/tex]So:(a) Find the number of passengers that will maximize the revenue received from the flight.We have to see if [tex]n_{v}[/tex] is higher than 80.We have that, for [tex]n > 80[/tex], [tex]R(n) = -5n^{2} + 986n[/tex], so [tex]a = -5, b = 986[/tex]The number of passengers that will maximize the revenue received from the flight is:[tex]n_{v} = -\frac{b}{2a} = -\frac{986}{2(-5)} = 98.6[/tex]Rounding up, the number of passengers that will maximize the revenue received from the flight is 99.(b) Find the maximum revenue.This is [tex]R(99)[/tex].[tex]R(n) = -5n^{2} + 986n[/tex][tex]R(99) = -5*(99)^{2} + 986*(99) = 48609[/tex]The maximum revenue is $48,609.