Question

a farmer is designing a rectangular area for his chickens. he has 240 feet of fencing that will form three sides of the rectangle; the fourth side will be formed by the barn and will not require any fencing. if he wants to maximize the area of the rectangle, what lengths should he use for the three sides formed by the fencing? 7 of 7 questions 80 ft, 80 ft, 80 ft 40 ft, 160 ft, 40 ft 60 ft, 120 ft, 60 ft 20 ft, 200 ft, 20 ft

Explanation:

Step1: Let the length parallel to the barn be \(x\) and the other two sides be \(y\)

We know that \(x + 2y=240\), so \(x = 240 - 2y\). The area \(A=xy=y(240 - 2y)=240y-2y^{2}\)

Step2: Find the maximum of the area function

For a quadratic function \(A(y)=- 2y^{2}+240y\) (in the form \(ay^{2}+by + c\) with \(a=-2\), \(b = 240\)), the vertex of the parabola \(y=-\frac{b}{2a}\)
Substitute \(a=-2\) and \(b = 240\) into \(y=-\frac{b}{2a}\), we get \(y=-\frac{240}{2\times(-2)} = 60\)
If \(y = 60\), then \(x=240-2\times60=120\)

Answer:

C. 60 ft, 120 ft, 60 ft