SOLUTION: The primeter of a rectangle is 30 cm, Find the whole-number dimensions of the rectangle with: (a) the greatest area (b) the least area

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The primeter of a rectangle is 30 cm, Find the whole-number dimensions of the rectangle with: (a) the greatest area (b) the least area      Log On


   

Question 588428: The primeter of a rectangle is 30 cm,
Find the whole-number dimensions of the rectangle with:
(a) the greatest area
(b) the least area
Found 2 solutions by solver91311, josmiceli:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


If you are allowed to use the fact that the largest area rectangle for a given perimeter is a square with side measure of the perimeter divided by 4, then divide 30 by 4 to get 7.5 and then round up to 8.

If you are required to prove that the largest area is a square then write back and I'll show you a proof.

The smallest would be a 0 by 15 rectangle with an area of zero, but a rectangle with a zero dimension is not a rectangle, it is a line. The restriction to stick with whole numbers means the two short sides must measure 1, leaving 14 for the two long sides. 1 times 14 is 14.

John

My calculator said it, I believe it, that settles it
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Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The formula for perimeter of a rectangle is
+P+=+2L+%2B+2W+
The formula for Area is
+A+=+L%2AW+
----------
given:
+P+=+30+ cm
+30+=+2L+%2B+2W+
+2W+=+30+-+2L+
+W+=+15+-+L+
Substitute this into A
+A+=+L%2A%28+15+-+L+%29+
+A+=+-L%5E2+%2B+15L+
Here's a plot of this equation.
L is on the horizontal axis
+graph%28+400%2C+400%2C+-4%2C+20%2C+-4%2C+60%2C+-x%5E2+%2B+15x+%29+
The value of L that gives a maximum A
is +-b%2F%282a%29+=+-15%2F%282%2A%28-1%29%29+
+-b%2F%282a%29+=+7.5+
The rectangle with maximum area is a
square with sides = 7.5 cm
+7.5%2A7.5+=+56.25+ cm2
The minimum area is 15 cm2