The length of a rectangle is 2 cm less than 6 times the width. find the possibilities for the width of the rectangle such that the area is not more than 840 cm2

To solve this problem, let us first assign the variables. Let us say that:

l = length of rectangle

w= width of rectangle

The given problem gives us the relation that:

l = 6 w – 2                            ---> 1

We know that the formula for area of rectangle is given as:

A = l w                                   ---> 2

Substituting equation 1 into 2:

A = (6 w – 2) w

A = 6 w^2 – 2 w

The area must not be greater 840 cm^2, therefore:

840 cm^2 = 6 w^2 – 2 w

w^2 – (1/3) w = 140

By completing the square:

w^2 – (1/3) w + (1/36) = 140 + (1/36)

(w – 1/6)^2 = 5041/36

w – 1/6 = ± 11.83

w = -11.66, 12

Therefore the maximum value that the width can take is 12 cm. Therefore the possibilities of width such that the area does not exceed 840 cm^2 is from 1 to 12 cm.

1 cm to 12 cm