Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 3 - Review Exercises - Page 300: 23b


When $p=40~dollars$ the maximum revenue is obtained: $R=16,000~dollars$.

Work Step by Step

We need to find the vertex of $R(p)=-10p^2+800p$ $R(p)=-10p^2+800p~~$ ($a=-10,b=800,c=0$): $-\frac{b}{2a}=-\frac{800}{2(-10)}=40$ $f(40)=-10(40)^2+800(40)=-16000+32000=16000$ Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(40,16000)$ That is, when $p=40~dollars$ the maximum revenue is obtained ($R=16,000~dollars$).
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