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Goshia [24]
3 years ago
5

A restaurant sells about 330 sandwiches each day at a price of $6 each $.25 decrease in price. 15 more sandwiches are sold per d

ay. How much should the restaurant charge to maximize daily revenue? Explain each step of your solution. What is the maximum daily revenue?
Mathematics
1 answer:
zmey [24]3 years ago
3 0

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 -3.75x^{2}

=1980 + 7.5x -3.75x^{2}

=1980-3.75 (-2x+x^{2}) ----> taking out common

now, to make a complete square lets add and subtract 1 inside the parentheses

=1980-3.75(-1+1-2x+x^{2})

=1980 +3.75 -3.75(x^{2} -2x +1)

=1983.75 -3.75 (x-1)^{2}---->(1)

as (x-1)^{2} is positive always, minimize the other term in order to maximize the total revenue.

so the minimum possible value of (x-1)^{2} = 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

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