Answer
maximum $f(20)= 1250$.
Work Step by Step
1. Given $f(x)=-3x^2+120x+50$, we have $a=-3\lt0, b=120$, thus it has a maximum.
2. We can find the maximum at $x=-\frac{b}{2a}=-\frac{120}{2(-3)}=20$ with $f(20)=-3(20)^2+120(20)+50=1250$.
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 6 - Analytic Trigonometry - Section 6.4 Trigonometric Identities - 6.4 Assess Your Understanding - Page 498: 113
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