Answer
The two numbers are $8$ and $-8.$
Minimum product: $-64$
Work Step by Step
See p.340: Strategy for Solving Problems lnvolving Maximizing or Minimizing Quadratic Functions
1. $\quad $Read the problem carefully and decide which quantity is to be maximized or minimized.
2. $\quad $Use the conditions of the problem to express the quantity as a function in one variable.
3. $\quad $Rewrite the function in the form $f(x)=ax^{2}+bx+c$.
4. $\quad $Calculate $-\displaystyle \frac{b}{2a}$.
If $a>0, f$ has a minimum of $f$($-\displaystyle \frac{b}{2a}$) at $x=-\displaystyle \frac{b}{2a}$.
If $a<0, f$ has a maximum of $f$($-\displaystyle \frac{b}{2a}$) at $x==-\displaystyle \frac{b}{2a}$
5. $\quad $Answer the question posed in the problem.
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1. Product is to be minimized
2. Let x be the greater of the numbers.
Then, the other number is $x-16$
Their product, $f(x)=x(x-16)$ is a quadratic function.
3.$\quad f(x)=x^{2}-16x$
$4.\qquad a=1, b=-16, c=0$
$-\displaystyle \frac{b}{2a}=-\frac{-16}{2(1)}=8$
5. $a > 0$,
The product is minimum when $x=8.$
The other number is $8-16=-8.$
The minimum product is $f(8)=8(-8)=-64$
