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Some word problems requiring the use of factorization?
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apriori
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#1
 Some word problems requiring the use of factorization?
I need to be able to solve some of these word problems, but I'm not sure how to start..

I will be VERY thankful if step-by-step instructions could be given
on how these types of word problems could be solved! I'm really confused
about how these can be translated into equations which could be
factorized.


1. A farmer plans to use 21m of fencing to enclose a rectangular pen
having an area of 55m. Only three sides of the pen need fencing because
part of the existing wall will form the fourth side. Find the dimensions of the pen.


2. A decorator plans to place a rug in a room 9m by 12m in such a way that
a uniform strip of flooring around the rug will remain uncovered. If the rug is to
cover half the floor space, what should the dimensions of the rug be?


3.A rancher plans to use 160yards of fencing to enclose a rectangular corral
and to divide it into two parts by a fence parallel to the shorter sides of the corral.
Find the dimensions of the corral if the area is 1000 yards.

PostPosted: Sat Nov 07, 2009 8:19 pm  Back to top 
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Stokes93
P versus NP
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#2
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1. Let the perimeter be p and the area be a: and let the sides on the rectangle by x and y. Then we know that p = 2y + x = 21 and a = xy = 55

Using the first equation you get x = 21 - 2y - plugging this back into the second equation you obtain 2y^2 -21y + 55 = 0 which can be factored as (2y - 11)(y-5) = 0 yielding y = 11/2 and y = 5. Then use these values of y to finish off the problem by solving for x - final solutions x = 11, y = 5 and x = 10, y = 11/2


PostPosted: Sat Nov 07, 2009 9:10 pm  Back to top 
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tornado.adv4
Poincare Conjecture
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#3
It helps to draw a picture for these.
2

unitsize(.5cm);
pair A,B,C,D,E,F,G,H,K,L,M,N;
A=(0,0);
B=(0,9);
C=(12,9);
D=(12,0);
E=(1.5,7.5);
F=(10.5,7.5);
G=(10.5,1.5);
...

The dimensions of the rug (in green) are (9-2x) by (12-2x).
Then the area of the rug is (9-2x)(12-2x).
This area must equal half of the floor space, or half of 9*12=108, which is 54.
So the equation is (9-2x)(12-2x)=54.
Solving,
\usepackage{amsmath}
\begin{align*}
(9-2x)(12-2x)&=54 \\
108-42x+4x^2&=54 \\
4x^2-42x+54&=0 \\
2x^2-21x+27&=0...
However, x=9 does not work because then you have no rug (or negative rug?).
So the only solution is x=3/2, which gives the dimensions of the rug to be \fbox{6\text{m\,by}\,9\text{m}}.



PostPosted: Sun Nov 08, 2009 5:08 pm  Back to top 
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modularmarc101
Navier-Stokes Equations
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#4
3
Let the rectangle have dimensions w and l, with l > w. Then we know that 2l + 3w = 160, and lw = 1000 \implies l = \frac {1000}{w}.

2 \cdot \frac {1000}{w} + 3w = 160

2000 + 3w^2 = 160w

3w^2 - 160w + 2000 = 0

(w - 20)(3w - 100) = 0

w = 20 or w = \frac {100}{3}

l = \frac {1000}{20} = 50 or l = \frac {1000}{\frac {100}{3}} = 30

We have that the dimensions (w,l) can be \boxed{(20, 50)}, or (\frac {100}{3}, 30). However, the second ordered pair gives a greater width than length; contradiction.

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PostPosted: Sun Nov 08, 2009 5:29 pm  Back to top 
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