Question 1035891
Let {{{ n }}} = the number of $1 increases in ticket price
Let {{{ r }}} = revenue from tickets
{{{ r = ( 1480 - 20n )*( 22 + 1*n )
{{{ r = 32560  - 440n + 1480n - 20n^2 }}}
{{{ r = -20n^2 + 1040n + 32560 }}}
{{{ r = -n^2 + 52n + 1628 }}}
The plot is a parabola with a maximum.
The {{{ n }}}-value of the maximum ios at:
{{{ n[max] = -b/(2a) }}} when the form is:
{{{ r = a*n^2 + b*n + c }}}
{{{ n[max] = -52 / ( 2*(-1)) }}}
{{{ n[max] = 26 }}}
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{{{ 22 + n = 22 + 26 }}}
{{{ 22 + n = 48 }}}
$48 / ticket gives maximum revenue
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{{{ r = ( 1480 - 20n )*( 22 + 1*n ) }}}
{{{ r = ( 1480 - 20*26 )*( 22 + 1*26 ) }}}
{{{ r = ( 1480 - 520 )*( 48 ) }}}
{{{ r = 960*48 }}}
{{{ r = 46080 }}}
The maximum revenue is $46,080 
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Here's the plot of revenue, {{{r}}},  and the
number of $1 increases in ticket price, {{{ n }}} 
{{{ graph( 400, 400, -10, 80, -6000, 60000, -20x^2 + 1040x + 32560 ) }}}