#### Answer

$8(8)=64$

#### 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 at $x=-\displaystyle \frac{b}{2a}$.
This minimum value is $f$($-\displaystyle \frac{b}{2a}$).
If $a<0, f$ has a maximum at $x=-2ab$.
This maximum value is $f$($-\displaystyle \frac{b}{2a}$)..
5. $\quad $Answer the question posed in the problem.
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1. Product is to be maximized
2. Let x be one of the numbers. Then, the other number is 16-x
Their product, $f(x)=x(16-x)$ is a quadratic function.
3.$\quad f(x)=16x-x^{2}$
$f(x)=-x^{2}+16x$
$4.\qquad a=-1, b=16, c=0$
$-\displaystyle \frac{b}{2a}=-\frac{16}{2(-1)}=8$
5. The product is maximum when $x=8.$
The other number is $16-8=8.$
The maximum product is f(8)$=8(8)=64$