Question

1. mr. coley has 2200 yards of fencing to enclose a rectangular area for his pet snail. express the area a of the rectangle as a function of the width w of the rectangle. for what value of w is the area largest? what is the maximum area?

Explanation:

Step1: Find the length in terms of width

The perimeter of a rectangle is \(P = 2L+2W\). Given \(P = 2200\), we have \(2L+2W=2200\), so \(L=\frac{2200 - 2W}{2}=1100 - W\).

Step2: Express the area as a function of width

The area of a rectangle is \(A = L\times W\). Substitute \(L = 1100 - W\) into the area formula: \(A(W)=(1100 - W)W=1100W - W^{2}\).

Step3: Find the value of \(W\) for maximum area

For a quadratic function \(y = ax^{2}+bx + c\) (\(a=- 1\), \(b = 1100\), \(c = 0\)), the vertex of the parabola \(x=-\frac{b}{2a}\).
Using the formula \(W=-\frac{1100}{2\times(-1)} = 550\).

Step4: Calculate the maximum area

Substitute \(W = 550\) into the area function \(A(W)\). \(A(550)=1100\times550-550^{2}=550\times(1100 - 550)=550\times550 = 302500\)

Answer:

The value of \(W\) for which the area is largest is \(550\) yards. The maximum area is \(302500\) square yards.