Question 1125541
Let {{{ r }}} = the total revenue
Let {{{ n }}} = the number of 5 cent increases in fare
{{{ r = ( 20000 + 2000n ) * ( 90 - 5n ) }}}
{{{ r = 2000*( 10 + n )*5*( 18 - n ) }}}
{{{ r = 10000*( 180 + 18n - 10n - n^2 ) }}}
{{{ r = 10000* ( -n^2 + 8n + 180 ) }}}
The 2nd term peaks at {{{ n[max] = -b/(2a) }}}
{{{ a.= -1 }}}
{{{ b = 8 }}}
{{{ n[max] = -8/(2*(-1)) }}}
{{{ n[max] = 4 }}}
The fare/rider that maximizes revenue is
{{{ 90 - 5n = 90 - 5*4 }}}
{{{ 90 - 5n =  70 }}}
70 cents/rider
———————-
Here’s a plot of r/10K and n:
{{{ graph( 400,400, -2, 20, -20, 250, -x^2 + 8x + 180 ) }}}