A piece of wire 24 cm long has the shape of a rectangle. Given that the width is w cm, show that the area, A cm^2....., of the rectangle is given...

If the length of the rectangle is 24 and the width is w,
then the width of the rectangle could be found from the formua of perimeter of the
rectangle.

The perimeter of a rectangle is the sum between
twice length and twice width.

P = 2*24 +
2*w

We'll factorize 2 and we'll
get:

P = 2(24+w)

The area of
the rectangle is:

A = 36 -
(6-w)^2

We'll expand the square and we'll
get:

36 - 36 + 12w - w^2

We'll
eliminate like terms and we'll get:

A = 12w -
w^2

We also know that the area of the rectangle is the
product of the length and the width.

24w = 12w -
w^2

We'll move all terms to one
side:

w^2 - 12w + 24w = 0

w^2
+ 12w = 0

We'll factorize by
w:

w(w+12)=0

Since the width
cannot be 0 or negative, the relation for area doesn't hold.