Answer
Length= $150$ ft and width$=300$ ft.
Maximum area: $45,000 $sq.ft
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.
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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.
