Question 242732: This is due in about... 1 hour (at 4oclock for me). I seriously need help or else my teacher will kill me :(
Two rectangles have a length and width that are consecutive even integers. The length and width of the larger rectangle are both two inches more than the length and width of the smaller rectangle. If the sum of the perimeters of the two rectangles is at most 64 inches, find the greatest possible perimeter of the larger rectangle. (Hint: Express all dimensions in terms of w, the width of the smaller rectangle.)
Found 3 solutions by Alan3354, scott8148, solver91311:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Two rectangles have a length and width that are consecutive even integers. The length and width of the larger rectangle are both two inches more than the length and width of the smaller rectangle. If the sum of the perimeters of the two rectangles is at most 64 inches, find the greatest possible perimeter of the larger rectangle. (Hint: Express all dimensions in terms of w, the width of the smaller rectangle.)
----------------
Smaller is w by w+2
Larger is w+2 by w+4
Perimeter = 2L + 2W
Total of 2 perimeters is 2(w + w+2) + 2(w+2 + w+4)
= 4w+4 + 4w+12
= 8w + 16
-----------
8w+16 <= 64
8w <= 48
w <= 6 inches
-------------
2(w+2 + w+4) = 36 inches around the larger rectangle
Answer by scott8148(6628)  (Show Source):
You can put this solution on YOUR website! smaller rectangle ___ width = w ; length = w+2 ; perimeter = 4w+4
larger rectangle ___ width = w+2 ; length = w+4 ; perimeter = 4w+12
8w + 16 ≤ 64
8w ≤ 48 ___ w ≤ 6
larger perimeter ___ 4w+12 ≤ 36
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
Come on now. I don't believe there is any jurisdiction in the world that prescribes capital punishment for the crime of ignorance of mathematics. Personally, I think there are cases where it would be a good idea, but that is another discussion for another time.
In order to make this work, you have to change the wording of the problem slightly, as follows:
Two rectangles have a length and width measured in inches that are consecutive even integers. The length and width of the larger rectangle are both two inches more than the length and width of the smaller rectangle. If the sum of the perimeters of the two rectangles is at most 64 inches, find the greatest possible perimeter of the larger rectangle.
Otherwise, the length and width could be in centimeters, or Angstroms, or whatever, and then you wouldn't be able to establish a fixed relationship between the smaller and larger rectangles.
Let  represent the width of the smaller rectangle in inches. If the length is the next consecutive even integer, then the length of the smaller rectangle must be  . The width of the larger rectangle is given as two inches more than the width of the smaller, so it is also . The length of the larger is then \ +\ 2\ =\ w\ +\ 4)
The perimeter of a rectangle is found by:
So now we can write a function for the perimeter of the smaller rectangle as a function of the width of the smaller rectangle:
\ =\ 2(w+2)\ +\ 2w\ =\ 4w\ + 4)
Using a similar process we can create the function for the perimeter of the larger rectangle, thus:
\ =\ 2(w+4)\ +\ 2(w+2)\ =\ 4w\ + 12)
Then the sum of the perimeters is:
\ +\ P_L(w)\ =\ (4w\ +\ 4)\ +\ (4w\ +\ 12)\ =\ 8w\ +\ 16)
Now we know that the sum of the perimeters must be at most 64 inches. That means that:
So the largest possible value of for which this last statement is true is 6.
Evaluate )
\ =\ 4(6)\ + 12\ =\ 36)
Check:
If  , then the smaller rectangle is 6 by 8 giving a perimeter of 12 + 18 = 28, and the larger rectangle must be 8 by 10 giving a perimeter of 16 + 20 = 36, and 28 + 36 = 64. Checks.
John
|
|
|
