Precalculus (6th Edition) Blitzer

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

Chapter 2 - Section 2.2 - Quadratic Functions - Exercise Set - Page 331: 64

Answer

The required pair of number is $\left( 12,-12 \right)$ , and their product is $\underline{-144}$.

Work Step by Step

Let us assume that the two numbers be x and y. Then, $x-y=24$ (I) Now, $\begin{align} & x-y=24 \\ & y=x-24 \\ \end{align}$ Calculate xy, $\begin{align} & xy=x\left( x-24 \right) \\ & xy={{x}^{2}}-24x \\ \end{align}$ Which is an upwards opening parabola, which attains its minimum value at $\frac{-b}{2a}$ , where $b=-24$ and $a=1$. $\begin{align} & x=\frac{-b}{2a} \\ & =\frac{-\left( -24 \right)}{2} \\ & =12 \end{align}$ Putting in the value of $x$ in the equation (I) we get: $\begin{align} & 12-y=24 \\ & -y=24-12 \\ & y=-12 \end{align}$ The minimum product is: $\begin{align} & xy=12\times \left( -12 \right) \\ & =-144 \end{align}$ Hence, the product is minimum when the numbers are 12 and -12 and the minimum product is $~-144$.
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