SOLUTION: A flat piece of wood with an area of 27 sq ft, and perimeter of 21 ft. How do I calculate the dimensions? I'm totally clueless. Thanks for our help.

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: A flat piece of wood with an area of 27 sq ft, and perimeter of 21 ft. How do I calculate the dimensions? I'm totally clueless. Thanks for our help.      Log On


   

Question 79468: A flat piece of wood with an area of 27 sq ft, and perimeter of 21 ft. How do I calculate the dimensions? I'm totally clueless. Thanks for our help.
Found 2 solutions by Nate, checkley75:
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
I don't know what shape the flat piece of wood is. I would assume it as a rectangle or a square.
lenght = l
width = w
2w + 2l = 21
and
lw = 27 or l = 27/w
Plug:
2w + 2l = 21
2w + 2(27/w) = 21
2w^2 + 54 = 21w
2w^2 - 21w + 54 = 0
2w^2 - 9w - 12w + 54 = 0
(2w^2 - 9w) + (-12w + 54) = 0
w(2w - 9) - 6(2w - 9) = 0
(w - 6)(2w - 9) = 0
w = 6
or
w = 9/2 = 4.5
The rest is explanatory.

Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
X*Y=27 X=27/Y
2X+2Y=21
2(27/Y)+2Y=21
54/Y+2Y=21
(54+2Y^2)/Y=21 CROSS MULTIPLY
54+2Y^2=21Y
2Y^2-21Y+54=0
(2Y-9)(Y-6)=0
2Y-9=0
2Y=9
Y=9/2
Y=4.5 ANSWER.
Y-6=0
Y=6 ANSWER.
PROOF
6*4.5=27
27=27
2*4.5+2*6=21
9+12=21
21=21