The annual total revenue for a product is given by

Upper R left parenthesis x right parenthesis equals 10 comma 000 x minus 2 x squaredR(x)=10,000x−2x2

dollars, where x is the number of units sold. To maximize revenue, how many units must besold? What is the maximum possible annual revenue?

in Algebra 1 Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

R(x)=10000x-2x² can be written R(x)=2(5000x-x²)=-2(x²-5000x+2500²-2500²).

R(x)=2(2500)²-2(x-2500)². When x=2500 the second term becomes zero and R(2500)=2(2500)² which is the maximum revenue. 2×2,500×2,500=2×6,250,000=12,500,000.

The maximum revenue is $12,500,000 when 2,500 units are sold.

The problem can also be solved through calculus:

Differentiate R(x): R'(x)=10000-4x=0 when the revenue is maximum. So 4x=10000, x=2500 units. And R(2500)=$12,500,000.

by Top Rated User (695k points)

Related questions

1 answer
3 answers
asked Apr 8, 2012 in Algebra 2 Answers by anonymous | 128 views
0 answers
Welcome to, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
84,070 questions
89,004 answers
6,767 users