Q:

The function f(x) = –(x – 3)2 + 9 can be used to represent the area of a rectangle with a perimeter of 12 units, as a function of the length of the rectangle, x. What is the maximum area of the rectangle?

Accepted Solution

A:
Answer:9. Step-by-step explanation:[tex]f(x) = -(x - 3)^2 + 9[/tex] is a parabola in its vertex form. For clarity, let [tex]f(x) = a (x - h)^2 + k[/tex] represent this function.[tex]a = -1[/tex].[tex]h = 3[/tex].[tex]k = 9[/tex].Note that [tex]a[/tex], the leading coefficient, is negative. Therefore, this parabola opens downwards. The vertex of the parabola would be [tex](h,\, k)[/tex], which in this question is the point [tex](3,\, 9)[/tex]. Since the parabola opens downwards, that vertex would be a local maximum (a crest) on its graph.Before concluding that the maximum area of this rectangle is [tex]9[/tex], make sure that [tex](3,\, 9)[/tex] is indeed on the graph of [tex]y = f(x)[/tex]. The length of a rectangle should be positive. Since [tex]x[/tex] represents the length of this rectangle, [tex]x > 0[/tex]. Also, since the perimeter should be less than [tex]12[/tex], the length of one side should be less than [tex]12 / 2 = 6[/tex]. Therefore, the domain of [tex]f[/tex] should be the open interval [tex](0,\, 6)[/tex]. (Endpoints not included.) Indeed, [tex]x = 3[/tex] is in that interval. [tex](3,\, 9)[/tex] would be on the graph [tex]y = f(x)[/tex]. Therefore, [tex]9[/tex] is indeed the maximum area of this rectangle.Side note: if the domain is a closed interval (i.e., endpoints included,) then consider checking the endpoints, as well.