Shakespear's Pizza sells 1 comma 100 large Vegi Pizzas per week for $18 a pizza. When the owner offers a $5 discount, the weekly sales increase to 1 comma 600. (a) Assume a linear relation between the weekly sales A(x) and the discount x. Find A(x). (b) Find the value of x that maximizes the weekly revenue. (c) Answer parts (a) and (b) if the price of one pizza is $9 and all other data are unchanged.

Explanation:

Solution:

A. Besides they intercept (0,100), the other data point is (5,600) since 600 pizzas are sold when there is a $5 discount. So, the slope of the line is

m = (600-100)÷(5-0)= 10

A(x) = 10x+ 100

B.

R = A(x)(Price) = (10x+ 100)(18-x)

R = 10(18-x) + (10x+ 100)(-1)

R = 80 -20x

If we set

R= 0 we obtain x = 80÷20= 4. Since R00 =-20 <0,

x gives the maximum of the function. So,The weekly revenue is maximized when x= $4

C.