Question

question: 15
a gardener has 120 feet of fencing. he wishes to enclose all four sides of a rectangular area with this fencing. a quadratic function, giving the area (a) of these rectangles, that models this problem situation is a(x) = 60x - x², where x represents the length of the rectangle. what is a reasonable domain and range for this area function?
a.
\tdomain = (0, 120)
\trange = (0, 60)

b.
\tdomain = (0, 30)
\trange = (0, 225)

c.
\tdomain = (0, 60)
\trange = (0, 900)

d.
\tdomain = (0, 120)
\trange = (0, 3600)

Explanation:

Step1: Analyze the domain

The perimeter of the rectangle is 120 feet. Let the length be \( x \) and the width be \( y \). Then \( 2(x + y)=120\), so \( x + y = 60\), and \( y=60 - x \). Since the length \( x>0 \) and the width \( y = 60 - x>0\), we have \( x>0 \) and \( 60 - x>0\), which implies \( 0 < x<60 \). So the domain of \( A(x)=60x - x^{2} \) is \( (0, 60) \).

Step2: Analyze the range

The function \( A(x)=60x - x^{2} \) is a quadratic function with \( a=- 1\), \( b = 60 \), \( c = 0 \). The vertex of a quadratic function \( ax^{2}+bx + c \) is at \( x=-\frac{b}{2a} \). So \( x=-\frac{60}{2\times(-1)} = 30 \). Substitute \( x = 30 \) into \( A(x) \): \( A(30)=60\times30-30^{2}=1800 - 900=900 \). Since \( a=-1<0 \), the parabola opens downwards, so the maximum value of \( A(x) \) is 900, and the minimum value (approaching) is 0 (when \( x \) approaches 0 or 60). So the range is \( (0, 900) \).

Answer:

C. domain = (0, 60)
range = (0, 900)