SOLUTION: What is the largest possible area of a rectangle with integer side length and a perimeter of 22?

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Question 142358This question is from textbook McDougal littell Math
: What is the largest possible area of a rectangle with integer side length and a perimeter of 22? This question is from textbook McDougal littell Math

Found 2 solutions by josmiceli, solver91311:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let one side = c
Let the other side= d
Let area = A
Let perimeter= P
A+=+cd
P+=+2c+%2B+2d
22+=+2c+%2B+2d
11+=+c+%2B+d
d+=+11+-+c
Now substitute into area formula
A+=+c%2A%2811+-+c%29
A+=+-c%5E2+%2B+11c
This is a parabola that has a peak. I know because
of the (-) sign in front of c%5E2
The peak is exactly between the x-intercepts, or at
c+=+%28-b%29%2F%282a%29 where a+=+-1 and b+=+11
because the equation is in the form A+=+ac%5E2+%2B+bc
c+=+%28-11%29%2F%28-2%29
c+=+11%2F2
P+=+2%2811%2F2%29+%2B+2d
22+=+11+%2B+2d
d+=+11%2F2
So, the max area is
A%5Bmax%5D+=+%2811%2F2%29%2811%2F2%29
A%5Bmax%5D+=+121%2F4
A%5Bmax%5D+=+30.25

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
P=2L+%2B+2W
A=LW

Given P = 22, so 2L+%2B+2W=22

A little algebra gets us: L=11-W

So A=LW=%2811-W%29W=-W%5E2%2B11W

This equation graphs to a convex down parabola with vertex at -11%2F%282%28-1%29%29=11%2F2, telling us that the maximum area is achieved when the length of the side is 11%2F2. The difficulty is that the problem asks for the maximum area of a rectangle with integer side lengths. The nearest integer smaller than 11%2F2 is 5 and the nearest integer larger than 11%2F2 is 6.

Substituting 5 for W in 2L+%2B+2W=22 gives us L=6, so obviously substituting 6 for W in 2L+%2B+2W=22 will give us L=5. Since 5 and 6 are the nearest integers to the side length that provides the maximum area (notice that all 4 sides would be 11%2F2 making a square) these must be the required side lengths and the maximum area for integer side lengths is 5%2A6=30