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.
Answer:
1 cm to 12 cm
