Question

david has 2000 yards of fencing to enclose a rectangular area. find the dimensions of the rectangle that maximize the enclosed area. what is the maximum area? a rectangle that maximizes the enclosed area has a length of □ yards and a width of □ yards.

Explanation:

Step1: Set up the equations

Let the length of the rectangle be \(l\) and the width be \(w\). The perimeter formula is \(P = 2l+2w\). Given \(P = 2000\), so \(2l + 2w=2000\), which simplifies to \(l + w=1000\), and \(l=1000 - w\). The area formula is \(A=l\times w\). Substitute \(l = 1000 - w\) into the area formula: \(A=(1000 - w)w=1000w-w^{2}\).

Step2: Find the maximum of the quadratic function

The area function \(A(w)=-w^{2}+1000w\) is a quadratic function of the form \(y = ax^{2}+bx + c\) (\(a=- 1\), \(b = 1000\), \(c = 0\)). The vertex of a quadratic function \(y = ax^{2}+bx + c\) has its \(x\) - coordinate (in our case \(w\)) given by \(w=-\frac{b}{2a}\).
Substitute \(a=-1\) and \(b = 1000\) into \(w=-\frac{b}{2a}\): \(w=-\frac{1000}{2\times(-1)} = 500\).
Since \(l=1000 - w\), when \(w = 500\), \(l=1000 - 500=500\).

Step3: Calculate the maximum area

Substitute \(l = 500\) and \(w = 500\) into the area formula \(A=l\times w\). \(A=500\times500 = 250000\) square yards.

Answer:

A rectangle that maximizes the enclosed area has a length of \(500\) yards and a width of \(500\) yards. The maximum area is \(250000\) square yards.