Question

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

Answer 1

Maximum area of the rectangle is $9cm^{2}$

Explanation:

Considering the dimensions to be in cm

$f(x) = -(x-3)^{2} +9\\f(x) = -(x^{2} +9 - 6x)+9\\f(x) = -x^{2} +6x\\f'(x) = -2x+6\\-2x+6 = 0\\2x=6\\x=3cm$

Putting the value of x = 3

$Perimeter = 2(x+b)\\12 = 2(3+b)\\6 = 3+b\\b= 3cm$

$Area of rectangle = x X b\\ = 3 X 3\\ = 9cm^{2}$

Therefore, maximum area of the rectangle is $9cm^{2}$