Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 4 Quadratic Functions and Factoring - 4.1 Graph Quadratic Functions in Standard Form - 4.1 Exercises - Problem Solving - Page 242: 55

Answer

They should increase the price by $0.75$.

Work Step by Step

$R(x)=(1+0.05x)(4000-80x)=-4x^2+120x+4000$ Let's compare $R(x)=-4x^2+120x+4000$ to $f(x)=ax^2+bx+c$. We can see that a=-4, b=120, c=4000. $a\lt0$, hence the graph opens down, and its vertex is a maximum. The maximum value is at $x=-\frac{b}{2a}=-\frac{120}{2\cdot(-4)}=15.$ Hence the maximum value is $f(15)=-4(15)^2+120(15)+4000=4900.$ Thus they should increase the price by $15\cdot0.05=0.75$.
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