During the 1950s the wholesale price for chicken for a country fell from 25¢ per pound to 14¢ per pound, while per capita chicken consumption rose from 23.5 pounds per year to 29 pounds per year. Assume that the demand for chicken depended linearly on the price. (a) Construct a linear demand function q(p), where p is in cents (e.g., use 14¢, not $0.14). Then, find the revenue function. R(p).

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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During the 1950s the wholesale price for chicken for a country fell from 25¢ per pound to 14¢ per pound, while per capita chicken
consumption rose from 23.5 pounds per year to 29 pounds per year. Assume that the demand for chicken depended linearly on
the price.
(a) Construct a linear demand function q(p), where p is in cents (e.g., use 14¢, not $0.14).
Then, find the revenue function. R(p).
R(p) =
(b) What wholesale price for chicken would have maximized revenues for poultry farmers?
¢ per pound
Second derivative test:
Your answer above is a critical point for the revenue function. To show it is a maximum, calculate the second derivative of the
revenue function.
R"(p)=
Evaluate R"(p) at your critical point. The result is ---Select---
which means that the revenue is
--Select---
v at the
critical point, and the critical point is a maximum.
Transcribed Image Text:During the 1950s the wholesale price for chicken for a country fell from 25¢ per pound to 14¢ per pound, while per capita chicken consumption rose from 23.5 pounds per year to 29 pounds per year. Assume that the demand for chicken depended linearly on the price. (a) Construct a linear demand function q(p), where p is in cents (e.g., use 14¢, not $0.14). Then, find the revenue function. R(p). R(p) = (b) What wholesale price for chicken would have maximized revenues for poultry farmers? ¢ per pound Second derivative test: Your answer above is a critical point for the revenue function. To show it is a maximum, calculate the second derivative of the revenue function. R"(p)= Evaluate R"(p) at your critical point. The result is ---Select--- which means that the revenue is --Select--- v at the critical point, and the critical point is a maximum.
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