Algebra and Trigonometry 10th Edition

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

Chapter 3 - 3.1 - Quadratic Functions and Models - 3.1 Exercises - Page 250: 72b


When $p=6.25~dollars$ the maximum revenue is obtained ($R=468.75~dollars$).

Work Step by Step

We need to find the vertex of $R(p)=-12p^2+150p$ $R(p)=-12p^2+150p~~$ ($a=-12,b=150,c=0$) $-\frac{b}{2a}=-\frac{-150}{2(-12)}=6.25$ $f(6.25)=-12(6.25)^2+150(6.25)=-468.75+937.50=468.75$ Vertex: $(-\frac{b}{2a},f(-\frac{b}{2a}))=(6.25,468.75)$ That is, when $p=6.25~dollars$ the maximum revenue is obtained ($R=468.75~dollars$).
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