Question

find the dimensions of a rectangle with a perimeter of 180 feet that has the maximum area. the side lengths are (square) feet. (use a comma to separate answers as needed.)

Explanation:

Step1: Set up the equations

Let the length of the rectangle be \(x\) and the width be \(y\). The perimeter formula is \(P = 2(x + y)\), and given \(P=180\), so \(2(x + y)=180\), which simplifies to \(x + y=90\), and \(y = 90 - x\). The area formula is \(A=xy\). Substitute \(y = 90 - x\) into the area formula: \(A(x)=x(90 - x)=90x - x^{2}\).

Step2: Find the maximum of the area function

Since \(A(x)=-x^{2}+90x\) is a quadratic function of the form \(y = ax^{2}+bx + c\) (\(a=-1\), \(b = 90\), \(c = 0\)). For a quadratic function \(y = ax^{2}+bx + c\), the vertex \(x\) - coordinate is given by \(x=-\frac{b}{2a}\).
Substitute \(a=-1\) and \(b = 90\) into \(x=-\frac{b}{2a}\), we get \(x=-\frac{90}{2\times(-1)} = 45\).
Since \(y=90 - x\), when \(x = 45\), \(y=45\).

Answer:

\(45,45\)