Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Cumulative Review Exercises - Page 516: 18

Answer

The required solution is $2\sec ,69ft$

Work Step by Step

$s\left( t \right)=-16{{t}^{2}}+64t+5$ And compare with $\begin{align} & a{{x}^{2}}+bx+c \\ & a=-16 \\ & b=64 \\ & c=5 \\ \end{align}$ Because $a<0$ Thus, the function's maximum value will occur at $\begin{align} & t=-\frac{b}{2a}=\frac{-64}{2\times \left( -16 \right)} \\ & t=2 \\ & s\left( 2 \right)=-16{{\left( 2 \right)}^{2}}+64\times 2+5 \\ & =69ft \\ & \end{align}$
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