Question

6.) A deli sells 640 sandwiches per day at a price of $8.00 each. A market survey shows

Answer 1

The demand function is

... q(p) = 640 + 40(8.00-p)/0.10 . . . a drop in price of 0.10 sells 40 more sandwiches

... q(p) = 3840 -400p . . . . . . . . . expanded form

... q(p) = 400(9.60 - p) . . . . . . . . factored form

The revenue is the product of price and the quantity sold.

... R(p) = p·q(p)

... R(p) = 400p(9.60 - p)

This is the equation of a quadratic function with zeros at p=0 and p=9.60. Since the zeros of a parabola are symmetric about its vertex, the vertex (maximum) will be halfway between those values, at p=4.80.

The deli should charge $4.80 for a sandwich to maximize revenue.

Answer 2

Given Revenue = No. of Sandwiches sold x price per sandwich

For every 10 cent drop, dell sells 40 more sandwhiches  

i.e. 1 cent drop, no. of sandwich +4

let x be how many of cent of price drop

No. of sandwiches sold = 640+4x when price = 8-0.01x

Revenue = (640+4x)*(8-0.01x)  

= 5120 - 6.4x +32x -0.04x^2  

= -0.04x^2 + 25.6x + 5120  

To find max revenue, differentiate by x and set it to 0

dRevenue/da = -0.08x + 25.6 = 0

0.08x = 25.6  

x = 25.6/0.08 = 320  

So the revenue is max when x=320 i.e. price = 8 - 0.01*320 = $4.8