Precalculus (6th Edition) Blitzer

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

Chapter 8 - Section 8.1 - Matrix Solutions to Linear Systems - Exercise Set - Page 895: 67


$60$ units, $7700$ dollars.

Work Step by Step

Step 1. Using the given function $y=ax^2+bx+c$ and the data from the table, we can set up the system of equations as $\begin{cases} a(30)^2+b(30)+c=5900 \\ a(50)^2+b(50)+c=7500 \\ a(100)^2+b(100)+c=4500 \end{cases}$ or $\begin{cases} 900a+30b+c=5900 \\ 2500a+50b+c=7500 \\ 10000a+100b+c=4500 \end{cases}$ Step 2. Taking the difference between the first and second, then the second and the third equations, we have $\begin{cases} 1600a+20b=1600 \\ 7500a+50b=-3000 \end{cases}$ or $\begin{cases} 80a+b=80 \\ 150a+b=-60 \end{cases}$ Step 3. Taking the difference between the two equations, we have $70a=-140$ and $a=-2$ Step 4. Back-substitute to get $b=80-80(-2)=240$ and $c=5900-900(-2)-30(240)=500$ and the function becomes $y=-2x^2+240x+500$ Step 5. The maximum of the function can be found at $x=-\frac{b}{2a}=-\frac{240}{2(-2)}=60$ with a maximum profit of $y=-2(60)^2+240(60)+500=7700$ dollars.
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