Precalculus (6th Edition) Blitzer

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

Chapter 2 - Mid-Chapter Check Point - Page 381: 27


For maximum profit, the company should manufacture $75$ cabinet per day and $\$1200$ is maximum profit.

Work Step by Step

The profit of the company is modelled as $p\left( x \right)=-{{x}^{2}}+150x-4425$ which is a quadratic equation with $a<0$. Thus, it would have a maximum value at $x=-\frac{b}{2a}$ $\begin{align} & x=-\frac{150}{2\left( -1 \right)} \\ & x=75 \end{align}$ $p\left( x \right)=-{{x}^{2}}+150x-4425$ Put $x=75$ in the above equation to maximize the company profit, $\begin{align} & p\left( 75 \right)=-{{\left( 75 \right)}^{2}}+150\left( 75 \right)-4425 \\ & =-5625+11250-4425 \\ & =-5625+6825 \\ & =1200 \end{align}$
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