the length of a rectangle is eight more than twice the width. if the perimeter is at most 132 inches, what is the greatest possible measure of its length?

Question

Let x= width Let 2x+8= length

Answer 1

The formula for the perimeter (P) of a rectangle is:  P = 2L + 2W      (L=length;  W=width)

 

The problem states that P≤ 132 in          ("at most")

   and   L = 8 + 2W

 

Now, the easy math:

    2W = L - 8             [from previous equation]

 

   P = 2L  +   (L - 8)          [from first formula]

 

   2L  +   (L - 8)   ≤ 132

       3L - 8 ≤ 132

        3L ≤ 140                 [keep the inequality when add to both sides]

        L ≤  46  2/3 

 

Checking:

   Is  P = 2(46.66) + 2(38.66/2) ) ≤  132    ?

               93.33  + 38.66  ≤  132   ?

                 132 ≤ 132   ? yes

 

Note: to use x=width, follow the same procedure with different values to get the same answer.

Answer 2

A rectangle has 4 sides, so we have to add 4 things together to get the perimeter. Two of them are x and two of them are 2x+8.

 

So we're adding (x) + (2x+8) + (x) + (2x+8)  (left side + top + bottom + right side).

 

So the total of the four sides is 6x + 16.

 

We want this total (the perimeter) to be no more than 132, so

 

6x + 16 ≤ 132

 

We solve an inequality like an equation, so subtract 16 from both sides:

 

6x ≤ 116

 

Divide both sides by 6.

 

x ≤ 19 1/3 inches, but we're not finished, because x represents the width of the rectangle. The length is 2x+6, so it is 2x(19 1/3)+6 = 

44 2/3 inches is the greatest length the rectangle can have.