## Precalculus (6th Edition) Blitzer

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