Question

what is the minimum product?

1. you have 100 yards of fencing to enclose a rectangular region. find the dimensions of the rectangle that maximize the enclosed area. what is the maximum area?

Explanation:

Step1: Define variables

Let the length of the rectangle be \( l \) and the width be \( w \). The perimeter of a rectangle is given by \( P = 2(l + w) \). We know the perimeter \( P = 100 \) yards, so \( 2(l + w)=100 \), which simplifies to \( l + w = 50 \), or \( l = 50 - w \).

Step2: Define the area function

The area \( A \) of a rectangle is \( A=l\times w \). Substituting \( l = 50 - w \) into the area formula, we get \( A(w)=(50 - w)w=50w - w^{2} \). This is a quadratic function in the form \( A(w)=-w^{2}+50w \), where \( a=- 1 \), \( b = 50 \), and \( c = 0 \).

Step3: Find the vertex of the parabola

For a quadratic function \( f(x)=ax^{2}+bx + c \), the \( x \)-coordinate of the vertex (which gives the maximum or minimum value) is at \( x=-\frac{b}{2a} \). Since \( a=-1<0 \), the parabola opens downwards, so the vertex is the maximum point. For our area function \( A(w)=-w^{2}+50w \), the width \( w \) that maximizes the area is \( w=-\frac{50}{2\times(-1)}=\frac{- 50}{-2}=25 \) yards.

Step4: Find the length and maximum area

If \( w = 25 \) yards, then \( l=50 - w=50 - 25 = 25 \) yards. So the rectangle is a square with side length 25 yards. The maximum area \( A=l\times w=25\times25 = 625 \) square yards.

Answer:

The dimensions of the rectangle that maximize the area are length \( 25 \) yards and width \( 25 \) yards (a square), and the maximum area is \( 625 \) square yards.