Question

1. a rectangular dog pen is constructed using a barn wall as one side and 60m of fencing for the other three sides. find the dimensions of the pen that give the greatest area and give the area.

Explanation:

Step1: Set up variables

Let the side parallel to the barn wall be \(x\) meters and the other two equal - length sides be \(y\) meters. Then the fencing equation is \(x + 2y=60\), so \(x = 60 - 2y\). The area formula \(A=xy\). Substitute \(x\) into the area formula: \(A=(60 - 2y)y=60y-2y^{2}\).

Step2: Find the maximum of the quadratic function

For a quadratic function \(A(y)=ay^{2}+by + c\) (here \(a=-2\), \(b = 60\), \(c = 0\)), the vertex of the parabola \(y=-\frac{b}{2a}\).
Substitute \(a=-2\) and \(b = 60\) into \(y =-\frac{b}{2a}\), we get \(y=-\frac{60}{2\times(-2)} = 15\).

Step3: Find the value of \(x\)

Substitute \(y = 15\) into \(x=60 - 2y\), then \(x=60-2\times15=30\).

Step4: Calculate the area

Substitute \(x = 30\) and \(y = 15\) into \(A=xy\), \(A=30\times15 = 450\).

Answer:

The dimensions of the pen are \(30m\) (parallel to the barn wall) and \(15m\) (perpendicular to the barn wall), and the maximum area is \(450m^{2}\).