Question

a rancher has 260 feet of fence with which to enclose three sides of a rectangular meadow (the fourth side is a river and will not require fencing). find the dimensions of the meadow with the largest possible area. (for the purpose of this problem, the width will be the smaller dimension (needing two sides); the length will be the longer dimension (needing one side).) length = feet width = feet what is the largest area possible for this meadow? area = feet - squared enter your answers as numbers. if necessary, round to the nearest hundredths.

Explanation:

Step1: Define variables

Let the width of the rectangle be $x$ (the side with two lengths of fencing) and the length be $y$ (the side parallel to the river). The total length of the fence is $2x + y=260$, so $y = 260 - 2x$.

Step2: Express the area function

The area $A$ of a rectangle is $A=xy$. Substitute $y = 260 - 2x$ into the area formula, we get $A(x)=x(260 - 2x)=260x-2x^{2}$.

Step3: Find the maximum of the area function

Since $A(x)$ is a quadratic - function of the form $y = ax^{2}+bx + c$ with $a=-2$, $b = 260$, and $c = 0$. The vertex of a quadratic function $y = ax^{2}+bx + c$ has its $x$ - coordinate at $x=-\frac{b}{2a}$.
For $A(x)=-2x^{2}+260x$, $x=-\frac{260}{2\times(-2)} = 65$.

Step4: Find the length

Substitute $x = 65$ into the equation $y = 260 - 2x$. Then $y=260-2\times65=130$.

Step5: Calculate the maximum area

Substitute $x = 65$ and $y = 130$ into the area formula $A = xy$. So $A=65\times130 = 8450$.

Answer:

length = 130 feet
width = 65 feet
area = 8450 feet - squared