SOLUTION: I know the formula for a rectangle P=2L+2W and A=LH, but what if the area and P are given? here is my question. Is it possible to have a rectangle with an area of 24 square met

Algebra ->  Rectangles -> SOLUTION: I know the formula for a rectangle P=2L+2W and A=LH, but what if the area and P are given? here is my question. Is it possible to have a rectangle with an area of 24 square met      Log On


   

Question 31402: I know the formula for a rectangle P=2L+2W and A=LH, but what if the
area and P are given? here is my question.
Is it possible to have a rectangle with an area of 24 square meters and a
perimeter of 96 meters? Give proof.
How do I use the formulas for this problem? Help.
Zack
Answer by Paul(988) About Me  (Show Source):
You can put this solution on YOUR website!
To see if this happens, lew x be the length and let y be the width:
2(x+y)=96
y=48-x (subsitution:)
AreA:
xy=24
Subsitute for y:
x(48-x)=24
48x-x^2=24
-x^2+48x-24=0
x^2-48x+24=0
Use the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%2848+%2B-+sqrt%28+%28-48%29%5E2-4%2A1%2A24+%29%29%2F%282%2A1%29+ --> REMOVE NEGATIVE SIGN
x+=+%2848+%2B+sqrt%28+2304-96+%29%29%2F%282%29+
x+=+%2848+%2B+sqrt%28+2208+%29%29%2F%282%29+
x=47
y=1
47*1=47
So it is not possible.
Paul.