The monthly revenue R (in thousands of dollars) from the sales of a digital picture frame is approximated by R(p) = −8p2 + 1360p, where p is the price per unit (in dollars). (a) Find the monthly revenues for unit prices of $50, $70, and $90. R(50) = $ thousand R(70) = $ thousand R(90) = $ thousand (b) Find the unit price that will yield a maximum monthly revenue. p = $ (c) What is the maximum monthly revenue? R = $ thousand
The monthly revenue R (in thousands of dollars) from the sales of a digital picture frame is approximated by R(p) = −8p2 + 1360p, where p is the price per unit (in dollars). (a) Find the monthly revenues for unit prices of $50, $70, and $90. R(50) = $ thousand R(70) = $ thousand R(90) = $ thousand (b) Find the unit price that will yield a maximum monthly revenue. p = $ (c) What is the maximum monthly revenue? R = $ thousand
Elementary Algebra
17th Edition
ISBN:9780998625713
Author:Lynn Marecek, MaryAnne Anthony-Smith
Publisher:Lynn Marecek, MaryAnne Anthony-Smith
Chapter8: Rational Expressions And Equations
Section8.9: Use Direct And Inverse Variation
Problem 501E: The force needed to break a board varies inversely with its length. If Tom uses 20 pounds of...
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The monthly revenue R (in thousands of dollars) from the sales of a digital picture frame is approximated by
R(p) = −8p2 + 1360p,
where p is the price per unit (in dollars).
(a) Find the monthly revenues for unit prices of $50, $70, and $90.
(b) Find the unit price that will yield a maximum monthly revenue.
(c) What is the maximum monthly revenue?
R(50) | = | $ thousand |
R(70) | = | $ thousand |
R(90) | = | $ thousand |
(b) Find the unit price that will yield a maximum monthly revenue.
p = $
(c) What is the maximum monthly revenue?
R = $ thousand
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