Question 1156061
Let {{{ R }}} = monthly revenue
Let {{{ n }}} = number of $1 decreases in price
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{{{ R = ( 30 - 1*n )*( 80 + 5n ) }}}
{{{ R = 2400 - 80n + 150n - 5n^2 }}}
{{{ R = -5n^2 + 70n + 2400 }}}
The n-value of the peak is at:
{{{ n[max] = -b/(2a) }}}
{{{ n[max] = -70/( 2*(-5)) }}}
{{{ n[max] = 7 }}}
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If the store lowers the price by $7, the 
monthly revenue will be maximum
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Here's the plot of {{{ R }}} and {{{ n }}}:
{{{ graph( 500, 500, -2, 20, -300, 3000, -5x^2 + 70x + 2400 ) }}}
The maximum revenue is:
{{{ R[max] = ( 30 - 1*7 )*( 80 + 5*7 ) }}}
{{{ R[max] = 23*115 }}}
{{{ R[max] = 2645 }}}
$2,645