## College Algebra (6th Edition)

$10(10)=100$
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 $f(10)=10(10)=100$