## College Algebra (6th Edition)

The two numbers are $8$ and $-8.$ Minimum product: $-64$
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 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 $f(8)=8(-8)=-64$