9 some dude wants two rectangular enclosures against his house for some reason. he has 300 ft of fencing…

Explanation:

Step1: Express the length in terms of (x)

Let the width of each enclosure be (x). There are (3) widths and (1) length. The total fencing is (300) ft. So, (3x + l=300), and (l = 300 - 3x).

Step2: Find the area equation

The area (A) of the two - enclosures (a rectangle) is (A=x\times l). Substitute (l = 300 - 3x) into the area formula. So, (A(x)=x(300 - 3x)=300x-3x^{2}).

Step3: Find the maximum of the quadratic function

The area function (A(x)=- 3x^{2}+300x) is a quadratic function of the form (y = ax^{2}+bx + c) ((a=-3), (b = 300), (c = 0)). The (x) - coordinate of the vertex of a quadratic function (y=ax^{2}+bx + c) is given by (x=-\frac{b}{2a}). Substitute (a=-3) and (b = 300) into (x=-\frac{b}{2a}): (x=-\frac{300}{2\times(-3)} = 50).

Step4: Calculate the maximum area

Substitute (x = 50) into the area function (A(x)=300x-3x^{2}). (A(50)=300\times50-3\times50^{2}=15000-3\times2500=15000 - 7500=7500) square feet.

Step5: Find the dimensions

We know (x = 50) ft (width). Substitute (x = 50) into (l=300 - 3x), (l=300-3\times50=300 - 150 = 150) ft (length).

Answer:

a) (A(x)=300x - 3x^{2}) b) (7500) square feet c) Width (x = 50) ft, length (l=150) ft