You have 24 feet of fencing that you are using to make a rectangular dog pen. You want the dog pen to enclose 150 square feet.
Is it possible for the 24 feet of fencing to enclose an area of 150 square feet? Explain.
P=24feet=LW length times width
A=150ft
2(150/W) +2W=24
W=????????
2007-04-13 07:57:27 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Your equations are wrong.
You should have P = 2L + 2w = 24, A = LW.
L+W = 12
LW = 150.
Let's solve this system and see if it has positive
real solutions. Solve the second equation for W
and substitute in the first one.
L + 150/L = 12
L² - 12L + 150 = 0.
But the discriminant of this quadratic is negative
since b² - 4ac = 144 -600 = -456.
So there are no real solutions
and the answer to your question is no.
2007-04-13 08:08:14 · answer #1 · answered by steiner1745 7 · 1⤊ 0⤋
Your perimeter and area formulas are off.
P = 2L + 2W
A = L * W
I don't know if you have learned this already or not, but a square is the most efficient use of a rectangle. The most area you can get in a rectangle of a set perimeter is where L = W, or a square.
So, with 24 feet of fencing, you can have L = W = 6 feet.
L * W = 6 * 6 = 36.
It certainly won't hold 150 square feet.
Even if you didn't know that the square maximizes the area of your rectangle, you can solve the following series of equations:
P = 2L + 2W
A = L * W
24 = 2L + 2W
150 = L * W
You know that L = 12 - W when you solve for L in the first equation. Plug that into your second equation to get:
150 = W (12 - W) = 12W - W^2 <-- Solve for W:
W^2 - 12W + 150 = 0
Using the quadratic formula, you see that W = 6 +/- SQRT(-456)/2. W is an imaginary number. You can tell this just by looking at the discriminant.
Since W must be an imaginary number, you know that you cannot solve this with real numbers. There is no way you can enclose 150 square feet with a rectangle made up of 24 square feet.
2007-04-13 08:02:16 · answer #2 · answered by Rev Kev 5 · 2⤊ 0⤋
P=24=2W+2L
A=150=WL<<
A=L(24-2L)/2=L(12-L)=12L-L^2
take derivative=12-2L ....set equal to zero
12=2L...L=6........Area is max at L=6
since L=6 put that in the first equation P=2W+2(6)=24...2W=12....W=6 and L=6 are your two numbers at which the area would be a maximum
6X6=36 its definitely not enough
2007-04-13 08:07:03 · answer #3 · answered by saeuta 3 · 1⤊ 0⤋
OK! The answer is ...no!
L * W = 150
and
2*L + 2*W = 24
2 equations in 2 variables. Solve!
2*L + 2*W = 24
L+W = 12
L= 12 - W
Substitute,
(12-W)*W = 150
12W - W^2 -150 = 0
W^2 - 12*W +150 = 0
Using the quadratic formula, you will find that there are no real solutions!
Please check the problem again.
2007-04-13 08:01:48 · answer #4 · answered by Jerry P 6 · 0⤊ 0⤋
umm.... no
The greatest area that you will be able to get is a square pen... The biggest square that you can get is a 6x6 pen... (24/4)
The greatest area attained from the pen is 36 ft
2007-04-13 08:02:54 · answer #5 · answered by Anonymous · 1⤊ 0⤋
do your own homework.
2007-04-13 08:05:47 · answer #6 · answered by Brutus Maxius 3 · 0⤊ 2⤋