Question 683794
the income will go up {{{linearly}}} for each passenger until the {{{100th}}} one, where the income would be ${{{60 * 100 = 6000}}}


after that, evidently, the next passenger would pay only ${{{59.50}}}, and the {{{100}}} other passengers would also pay ${{{0.50}}} less

let's put that in equation:

{{{Income = (60 -(n-100)/2)*n}}}

where "{{{n}}}" is the number of passenger

note that this equation applies only for "{{{n}}}" {{{greater}}} or {{{equal}}} to {{{100}}}

the {{{(n-100)/2}}} is the bit that takes away {{{50}}} cents for each passenger above the {{{100th}}}; and the "{{{n}}}" is the number of passenger

A bit of algebra turns the income equation to

{{{income=110n -n^2/2 }}}

Now, we want to maximize the income. Calculus to the rescue, as the local maximum implies that the derivative of the equation equals zero.

The derivative of the equation is

{{{110 - n}}}

If we put this equal to zero, then {{{n}}} has to be {{{110}}}.

That is thus the number of passengers that {{{maximizes}}} income.

Plugging {{{n=110}}} into the income equation means that said income would be 
${{{6050}}}.

And we can easily verify that this is indeed the {{{maximum}}}, by seeing what the income would be for {{{109}}} passengers (it would be ${{{6049.50}}}) and {{{111}}} passengers (${{{6049.50}}} also).