Question

A pizza parlor has been experimenting with lowering the price of their large one-topping pizza to promote sales. The average revenues from the sale of large one-topping pizzas on a Friday night (5 P.M. to midnight) at various prices are given below. Revenue from the sale of Pizzas at Different Prices Price Revenue (dollars x) (dollars R) 9.25 1202.50 1228.50 10.50 11.75 1210.25 13.00 1131.00 1054.50 14.25 9.25 SXS 14.25. (Round all numerical values to two (a) Find the function for the quadratic model that gives the average revenue in dollars where x is the price in dollars of a large one-topping pizza, with data from decimal places.) R(x) = dollars (b) Calculate the rate of change of revenue at a price of $9.25 and at a price of $11.50. (Round your answers to the nearest cent.) $9.255 per dollar $11.50 $ per dollar (c) Use the model to calculate the change in revenue if the price is increased from $9.25 to $10.25 and from $11.50 to $12.50. (Round your answers to the nearest cent.) $9.25 to $10.25 $ $11.50 to $12.50 $ (d) Explain why the approximate change is an overestimate of the change in price from $9.25 to $10.25 but an underestimate of the change in price from $11.50 to $12.50. The marginal price is found using a linearization of the revenue function. The revenue function is --Select- with a G-Select-- value. When R'x) > 0, the linearization because the tangent is --Select- R(x). When R'(x) < 0, the linearization --Select the change because the tangent is --Select--- R(x). --Select the change

Answer 1

given: Revenue from the sale of pizza at • different prices Price (dollars, Revenue 9.25 I 1202.50 10.50 12 28.50 12 10.25 11.75 13.00 1131.00 14.25 losu.so (a). in order to find the function for the quadratic model for the average revenue in dollars where x is the price in dollars, 9.255 25 1925) We will use T1-83 software calculator (1) first press 'STAT button and then click & EDIT (ii). Fill the table entries (in). Press 'STAT. CLICK SCALE (iv). Press en enter By the above steps, we will get the following Oudratic models: - - 12.168² +254.28x - 10s. 6 RO). - 12.16x² + 254.28x- 105.6 Ang

(b) rate of change of revenue at x-39.253.2 x=$11.50 3 1 we know that rate of change of revenue is r(x)= I CR(x)) da r(x)= d (-12.168² +254.28x-105-6) dla at r(x) = -12.16 (22) + 254.28 (1) -0 r(x)= - 24.320 + 254.28 - 9.25 r(9.25) = -24.32 (9.25) + 254.28 = $99.32 x= 1.50 (11.ro) -- 24:32 (11:00) + 254.28 -- $25.40 at hence Ir (9.25) = $29.327 Ir (11.50) – $25.40 Ang. (e) Change in a ich x: $9.25 $10.25 & 2: $11.50 -)$12.10 = R(10.25)- R19.25) - t - 12.16 (10. 25) ²+254.28(10.25) - 105.6 - (- (12.16) (g. 25)² + 254.28 - 1056) - 1223.21-12 06.05 - 17.16 & R 110.25). R (9.25) = $17.16 Any

(1) R (12.50) - R (11.50): - 12.16 (12:10) 2 254.28 (12.50) – 105.6 - (-12-16 (110)? +254.28 (1/10) - 1056)) = 1172.90-1210.46 = -37.56 TR (12.50) R (1100) = – $37.56)
hence

a. R(x) = -12.16x^2+254.28x-105.6

b. Rate of change

r(9.25) = $29.32

r(11.50) =-$25.40

c. R(10.25)-R(9.25)= $17.16 and

R(12.50)-R(11.50)= -$37.56

d. The marginal revenue is found using a linearization of R(x). R(x) is concave down with maximum value. When r(x)>0, linearization overestimates the change because the tangent is abouve R(x). Thus the approximate change in an oversrimate of the change in price from $9.50 to $10.25.

Thank you!