Derivative of the (sqrt of t/4t-3)
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Answer:
restaurant should charge $(6-0.25) = $5.75per sandwich to maximize daily revenue.
the revenue is $1983.75
Step-by-step explanation:
to calculate current revenue= $6 x 330 = $1980
suppose x as the number of times the price to be dropped by $0.25
then find new price.. i.e
new price= $(6-0.25x)
and, new sell=330 +15x sandwiches
therefore, the new revenue would be= (6-0.25x)(330 +15x)
in order to maximize the current revenue, simplify the above equation and make it complete square using x
(6-0.25x)(330 +15x)
=1980-82.5x +90x -3.75
=1980 + 7.5x -3.75
=1980-3.75 (-2x+) ----> taking out common
now, to make a complete square lets add and subtract 1 inside the parentheses
=1980-3.75(-1+1-2x+)
=1980 +3.75 -3.75( -2x +1)
=1983.75 -3.75 ---->(1)
as is positive always, minimize the other term in order to maximize the total revenue.
so the minimum possible value of = 0
therefore, x=1
putting x in eq(1) the revenure becomes,
$(1983.75-0)=> $1983.75
therefore, restaurant should charge $(6-0.25) = $5.75per sandwich to maximize daily revenue.
the revenue is $1983.75