Shakespear's Pizza sells 1 comma 100 large Vegi Pizzas per week for $16 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 $8 and all other data are unchanged.

(a) A(x) = 100x + 1100

(b) x = 2.5 USD

(c) A(x) = 100x + 1100 and x = -1.5 USD

Step-by-step explanation:

(a) Since there's a linear relation between weekly sales A(x) and the discount, we can model it as the following expression

A(x) = mx + b

where x is the discount, m is the slope and b is the intersect of the linear equation. We can solve for slope first

To solve for b, we can just plug in m = 100, x = 0 and A(0) = 1100

1100 = 100*0 + b

b = 1100

Therefore A(x) = 100x + 1100

With discount x, the actual price would then be p - x. Where p is the original undiscounted pricing ($16 in this case)

And with sale A(x), the revenue would then be

To find the maximum value of this, we can take the 1st derivative and set it to 0

R' = -200x + 100p - 1100 = 0

200x = 100p - 1100

when p = $16 then x = 0.5*16 - 5.5 = 8 - 5.5 = 2.5 USD

(c) when the price p is $8 then A(x) would sill be 100x + 1100 because it doesn't depend on p, but the discount. But to maximize the revenue, x will need to be

0.5p - 5.5 = 0.5*8 - 5.5 = 4 - 5.5 = -1.5 USD

So the Pizza owner would need to raise the price by 1.5 USD because originally the pizza is already inexpensive.