Answer
The two numbers are $12$ and $-12.$
Minimum product: $-144$
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 $x-24$
Their product, $f(x)=x(x-24)$ is a quadratic function.
3.$\quad f(x)=x^{2}-24x$
$4.\qquad a=1, b=-24, c=0$
$-\displaystyle \frac{b}{2a}=-\frac{-24}{2(1)}=12$
5. $a > 0,$
The product is minimum when $x=12$
The other number is $12-24=-12.$
The minimum product is $f(12)=12(-12)=-144$
