Question 1147418
Let {{{ k }}} = the number of $1 increases in price/ticket
{{{ R(p) = p*( 8000 - 500k ) }}}
{{{ -500p^2 + 13000p = p*( 8000 - 500k ) }}}
{{{ -500p + 13000 = 8000 - 500k }}}
{{{ -500p = -500k - 5000 }}}
{{{ p = k + 10 }}}
This shows the equation works
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{{{ R(p) = -500p^2 + 13000p }}}
The maximum is at {{{ p = -b/2a }}} when the equation has the form
{{{ R(p) = a*p^2 + b*p }}}
{{{ p[max]}}}  is at {{{ -13000/( 2*(-500)) }}}
{{{ p[max] = 13 }}}
The maximum revenue is at a ticket price of $13/ticket
which is after three $1 increases in price/ticket {{{ k=3 }}}
Here's the plot:
 {{{ graph( 400, 400, -5, 30, -10000, 100000, -500x^2 + 13000x) }}}