Question

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?

Answer 1

Answer:

The maximum area of rectangle is:

                   9 square units

Step-by-step explanation:

The function f(x) is given by:

$f(x)=-(x-3)^2+9$

where f(x) represent the area of rectangle.

Also, the area is maximized at the value of x where the derivative is equal to zero.

Hence,

$f'(x)=0\\\\i.e.\\\\-2(x-3)=0\\\\i.e.\\\\x=3$

Hence, when x=3 the area of the rectangle is maximum.

Also, the maximum area of rectangle is:

$f(3)=-(3-3)^2+9\\\\i.e.\\\\f(3)=0+9\\\\i.e.\\\\f(3)=9\ square\ units$

Answer 2

Answer:

Step-by-step explanation:

f(x) = –(x – 3)2 + 9

      = 3² - (x – 3)2  identity a²-b² = (a+b)(a - b)  where : a =3 and b = x-3

     = ( 3+x-3) (3-x+3)

    f(x) = x( 6-x)

the perimeter is ; 2(x +6-x) =12

the maximum area of the rectangle is : f(3) = 9 m²