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

Correct response:

• The maximum area of the rectangle is 9 square units

Methods used to find the maximum area

The given function for the area of the rectangle is f(x) = -(x - 3)² + 9

The perimeter of the rectangle, P = 12 units

The length of the rectangle = x

Required:

To find the maximum area of the rectangle.

Solution:

• The given function is given in vertex form of a quadratic equation which is; f(x) = a·(x - h)² + k

Where;

(h, k) = The vertex of the equation

By comparison, we have;

a = -1

h = 3 = The value of x at the vertex

k = 9 = The value of f(x) at the vertex = The maximum area

Therefore, the vertex, (h, k) = (3, 9)

Where;

h = 3 = The value of x that gives the maximum area

When the length, x = 3, the width of the rectangle also equals 3, and the rectangle is then a square

k = 9 = The maximum area of the rectangle

• The maximum area of the rectangle = 9 square units

Learn more about the vertex form of a quadratic equation here:

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