SOLUTION: How do you find the dimensions of a rectangle if it's perimeter is 170 feet and its area is 2800 square feet.

Algebra ->  Rectangles -> SOLUTION: How do you find the dimensions of a rectangle if it's perimeter is 170 feet and its area is 2800 square feet.       Log On


   

Question 848163: How do you find the dimensions of a rectangle if it's perimeter is 170 feet and its area is 2800 square feet.
Found 2 solutions by richwmiller, swincher4391:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
2L+2W=170
L+W=85
L=85-W
L*W=2800
(85-W)*W=2800
85W-W^2=2800
No such rectangle in the real world
It can't be.




Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
170 = 2L + 2W
2800 = L*W
2800/W = L
170 = 2*(2800/W) + 2W
170 = 5600/W + 2W
170W/W = 5600/W + 2W^2/W
W isn't 0, so we can ignore the denominator.
170W = 5600 + 2W^2
2W^2 - 170W +5600
2(W^2 - 85W + 2800)
This has no real solution. You can compute this by taking b^2 - 4ac, which is clearly less than 0.
Another thing we can notice that confirms this is say our rectangle has the same length and width. This maximizes our area (also known as a square). Then each side would have 42.5, and so 42.5^2 = 1806.25 which is the maximum area possible. This is not necessary to know, but it confirms that there is no real solution.