SOLUTION: A dog trainer has 100 ft of fencing that will be used to create a rectangular work area for dogs. If the trainer wants to enclose an area of 504 ft2, what will be the dimensions of

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A dog trainer has 100 ft of fencing that will be used to create a rectangular work area for dogs. If the trainer wants to enclose an area of 504 ft2, what will be the dimensions of      Log On


   

Question 1202100: A dog trainer has 100 ft of fencing that will be used to create a rectangular work area for dogs. If the trainer wants to enclose an area of 504 ft2, what will be the dimensions of the work area?
Found 3 solutions by josgarithmetic, math_tutor2020, ikleyn:
Answer by josgarithmetic(39620) About Me  (Show Source):
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Use all the 100 feet fence material?

w width
L length
system%28wL=504%2Cw%2BL=100%2F2%29

system%28wL=504%2Cw%2BL=50%29

w%2850-w%29=504
-
50w-w%5E2=504
w%5E2-50w=-504
highlight_green%28w%5E2-50w%2B504=0%29

-
discriminant 50%5E2-4%2A504=2500-4%2A504=2500-2016=484
22%5E2

w=%2850%2B-+22%29%2F2

w=14 OR w=36

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Dimensions be 14 feet by 36 feet.
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Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: 14 ft by 36 ft


Work Shown:

x = width
y = height

P = perimeter of the rectangle
P = 2*(width + height)
P = 2*(x + y)
Replace P with 100 as this is the perimeter, aka amount of fencing
100 = 2(x + y)
100/2 = x + y
50 = x + y
y = 50-x

A = area of the rectangle
A = width*height
A = x*y
A = x*(50-x)
A = 50x-x^2
A = -x^2+50x
Plug in A = 504 which is the desired area.
504 = -x^2+50x
0 = -x^2+50x-504
-x^2+50x-504 = 0
-(x^2-50x+504) = 0
x^2-50x+504 = 0

You could take a trial-and-error approach to factoring, but I think the quadratic formula is more efficient.
Plug in a = 1, b = -50, c = 504
x+=+%28-b+%2B-+sqrt%28b%5E2+-+4ac%29%29%2F%282a%29

x+=+%28-%28-50%29+%2B-+sqrt%28%28-50%29%5E2+-+4%281%29%28504%29%29%29%2F%282%281%29%29

x+=+%2850+%2B-+sqrt%282500-2016%29%29%2F%282%281%29%29

x+=+%2850+%2B-+sqrt%28484%29%29%2F%282%29

x+=+%2850+%2B-+22%29%2F%282%29

x+=+%2850+%2B+22%29%2F%282%29+ or +x+=+%2850+-+22%29%2F%282%29

x+=+%2872%29%2F%282%29+ or +x+=+%2828%29%2F%282%29

x+=+36+ or +x+=+14

If x = 36, then y = 50-x = 50-36 = 14
If x = 14, then y = 50-x = 50-14 = 36

The rectangle is 14 ft by 36 ft

Check:
Perimeter = 2*(14+36) = 2*50 = 100
Area = 14*36 = 504
The answers are confirmed.

Answer by ikleyn(52799) About Me  (Show Source):
You can put this solution on YOUR website!
.
A dog trainer has 100 ft of fencing that will be used to create a rectangular work area for dogs.
If the trainer wants to enclose an area of 504 ft2, what will be the dimensions of the work area?
~~~~~~~~~~~~~~~~~ 

L + W should be 100/2 = 50 ft.

So, if x is the width (in feet), then the length is (50-x) ft.


The area equation is

    x*(50-x) = 504

    50x - x^2 = 504

    x^2 - 50x + 504 = 0


Solve using the quadratic formula

    x%5B1%2C2%5D = %2850+%2B-+sqrt%28%28-50%29%5E2+-+4%2A1%2A504%29%29%2F2 = %2850+%2B-+22%29%2F2.


The roots are  x%5B1%5D = %2850+%2B+22%29%2F2 = 36  and  %2850+-+22%29%2F2 = 14.


They produce the same rectangle with dimensions 36 ft and 14 ft.    


ANSWER.  The dimensions of the rectangle are 36 ft by 14 ft.

Solved.