## College Algebra (6th Edition)

Length= $150$ ft and width$=300$ ft. Maximum area: $45,000$sq.ft
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. Rectangle area ( length $\times$ width ) is to be maximized 2. Let $x$ be the length of the rectangle. Then, the width y, is such that $y+2x=600$ (fence only one width, not the one along the river) That is, the width equals $600-2x$ Their product, the area, $f(x)=x(600-2x)$ is a quadratic function. 3.$\quad f(x)=600x-2x^{2}=-2x^{2}+600x$ $4.\qquad a=-2, b=600, c=0$ $-\displaystyle \frac{b}{2a}=-\frac{600}{2(-2)}=150$ 5. $a < 0,$ The area is maximum when length= $x=150$ ft The width is $600-2(150)=300$ ft The maximum area is $f(150)=150(300)=45,000$ sq.ft.