Question

A restaurant sells 330 sandwiches each day. For each $0.25 decrease in price, the restaurant sells about 15 more sandwiches. How much should the restaurant charge to maximize daily revenue? What is the maximum daily revenue? Each sandwich is $6 beforehand...

Answer 1

Answer:

The maximum daily revenue is $1983.75$ with 1 decrease in price.

Step-by-step explanation:

Let x be the number of $0.25 decreases in price and R to be the daily revenue in dollar.

Since, $Revenue= Number of price \cdot number of sold$

R(x)=$(6-0.25x)(330+15x)$

=$1980+90x-82.5x-3.75x^2$

= 1980+7.5x-3.75x^2

∴  $R(x)= -3.75x^2+7.5x+1980$     ....(1)

For a quadratic equation in standard form $y=ax^2+bx+c$, the axis of symmetry is a vertical line $x=\frac{-b}{2a}$

Here, a= -3.75 b=7.5

then,

$x=\frac{-b}{2a}=\frac{-7.5}{2\cdot (-3.75)}$$=\frac{7.5}{7.5} =1$

Substitute the value of x=1 in equation (1), to find the maximum revenue;

$R(1)=-3.75(1)^2+7.5(1)+1980$$=3.75+1980$=$\ 1983.75$

Therefore, the maximum daily revenue is, $\ 1983.75$ with x=1 decrease in price.

Answer 2

The restaurant should charge $ 5.75 per sandwich to maximize the revenue, which would be $ 1,983.25 per day.

Given that a restaurant sells 330 sandwiches each day, and for each $ 0.25 decrease in price, the restaurant sells about 15 more sandwiches, knowing that each sandwich costs $ 6, to determine how much the restaurant should charge to maximize daily revenue and what is the maximum daily revenue the following calculations must be performed:

• 330 x 6 = 1980
• 345 x 5.75 = 1983.75
• 360 x 5.5 = 1980

Therefore, the restaurant should charge $ 5.75 per sandwich to maximize the revenue, which would be $ 1,983.25 per day.

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