Question 960240
Let {{{ R }}} = average weekly revenue
Let {{{ n }}} = number of $5 reductions in the price of shoes
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(1)
{{{ R = ( 45 + 4n )*( 90 - 5n ) }}}
{{{ R = 4050 + 360n - 225n - 20n^2 }}}
{{{ R = -20n^2 + 135n + 4050 }}}
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(2)
Current weekly revenue is {{{ 45*90 = 4050 }}}
I need values for {{{ n }}} for which {{{ R >= 4050 }}}
{{{ 4050 = -20n^2 + 135n + 4050 }}}
{{{ 0 = -20n^2 + 135n }}}
{{{ 0 = 5n*( -4n + 27 ) }}}
The solutions are:
{{{ n = 0 }}}
{{{ -4n + 27 = 0 }}}
{{{ 4n = 27 }}}
{{{ n = 6.75 }}}
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The price per shoe when {{{ n = 6.75 }}} is
{{{ 90 - 5n = 90 - 5*6.75 }}}
{{{ 90 - 5n = 90 - 33.75 }}}
{{{ 90 - 5n = 56.25 }}}
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The range of prices to maintain or increase is
$56.25 to $90
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check:
{{{ R = ( 45 + 4*6.75 )*( 90 - 5*6.75 ) }}}
{{{ R = ( 45 + 27 )*56.25 }}}
{{{ R = 72*56.25 }}}
{{{ R = 4050 }}}
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(3)
I want the maximum possible {{{ R }}}
This is where {{{ n[max] = -b/(2a) }}}
{{{ a = -20 }}}
{{{ b = 135 }}}
{{{ n[max] = -135 / ( 2*(-20)) }}}
{{{ n[max] = 3.375 }}}
Plug this back in and find {{{ R[max] }}}
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{{{ R[max] = ( 45 + 4*3.375 )*( 90 - 5*3.375 ) }}}
{{{ R[max] = ( 45 + 13.5 )*( 90 - 16.875 ) }}}
{{{ R[max] = 58.5*73.125 }}}
{{{ R[max] = 4277.81 }}}
The maximum amount by which the store can 
increase its revenue from these shoes each week is
$4,277.81
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Here's the plot of the {{{ R }}} function:
{{{ graph( 500, 500, -4, 20, -200, 4700, -20x^2 + 135x + 4050 ) }}}