#### Answer

$10(10)=100$

#### 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.
----------------------------
1. Product is to be maximized
2. Let x be one of the numbers.
Then, the other number is $20-x$
Their product, $f(x)=x(20-x)$ is a quadratic function.
3.$\quad f(x)=20x-x^{2}$
$f(x)=-x^{2}+20x$
$4.\qquad a=-1, b=20, c=0$
$-\displaystyle \frac{b}{2a}=-\frac{20}{2(-1)}=10$
5.
The product is maximum when $x=10.$
The other number is $20-10=10.$
The maximum product is $f(10)=10(10)=100$