Question

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.

Answer 1

Answer:

(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

$m = \frac{s_2 - s_1}{x_2 - x_1} = \frac{1600 - 1100}{5 - 0} = \frac{500}{5} = 100$

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

$R = (p - x)A(x) = (p - x)*(100x + 1100)$

$R = 100xp + 1100p - 100x^2 - 1100x$

$R = -100x^2 + (100p - 1100)x + 1100p$

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

$x = \frac{100p - 1100}{200} = 0.5p - 5.5$

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.