SOLUTION: Cynthia Besch wants to buy a rug for a room that is 27ft wide and 35ft long. She wants to leave a uniform strip of floor around the rug. She can afford to buy 609 square feet of

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Question 1204421: Cynthia Besch wants to buy a rug for a room that is 27ft wide and 35ft long. She wants to leave a uniform strip of floor around the rug. She can afford to buy 609 square feet of carpeting. What dimensions should the rug​ have?
Found 6 solutions by ikleyn, math_tutor2020, josgarithmetic, MathLover1, Edwin McCravy, greenestamps:
Answer by ikleyn(52777) About Me  (Show Source):
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.
Cynthia Besch wants to buy a rug for a room that is 27ft wide and 35ft long.
She wants to leave a uniform strip of floor around the rug.
She can afford to buy 609 square feet of carpeting. What dimensions should the rug​ have?
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Let x be an unknown uniform width of the strip around the rug.


Then the dimensions of the rug are (27-2x) ft and (35-2x) ft.

The area equation is

    (27-2x)*(35-2x) = 609 sq. ft, the affordable area.


Simplify and find x

    27*35 - 70x - 54x + 4x^2 = 609

    4x^2 - 124x + 336 = 0

     x^2 - 31x + 84 = 0


Solve using the quadratic formula

    x%5B1%2C2%5D = %2831+%2B-+sqrt%2831%5E2+-+4%2A84%29%29%2F2 = %2831+%2B-+25%29%2F2

    x%5B1%5D = %2831+%2B+25%29%2F2 = 28;  x%5B2%5D = %2831+-+25%29%2F2 = 3.


First root  x%5B1%5D = 28 ft is too large value, and we deny it;  second root x%5B2%5D = 3 is good: we accept it.


ANSWER.  The dimensions of the rug should be  27-2*3 = 21 ft  and  35-2*3 = 29 ft.

Solved.



Answer by math_tutor2020(3816) About Me  (Show Source):
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A question like this has been asked before.

Review this link to see an example how to solve.
https://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.1197474.html

If you are still stuck, then read on.

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Draw out a diagram as indicated in the link above.

There will be a large rectangle that is 27 feet by 35 feet.
This represents the dimensions of the room.

Inside that larger rectangle will be a smaller rectangle with dimensions (27-2x) feet by (35-2x) feet.
x represents the width of the uniform strip.
This is the buffer gap between the edge of the rug to the edge of the wall.

The area of the smaller rectangle is (27-2x)(35-2x).
Set this equal to 609 because this is the amount of carpet she can buy.

(27-2x)(35-2x) = 609
27(35-2x)-2x(35-2x) = 609
945-54x-70x+4x^2 = 609
945-54x-70x+4x^2 - 609 = 0
4x^2 - 124x + 336 = 0
4(x^2 - 31x + 84) = 0
x^2 - 31x + 84 = 0

We could factor, but I think the quadratic formula is the most efficient pathway.
This is because we don't have to worry about guess-and-check.

Plug a = 1, b = -31, c = 84 into the quadratic formula below.
x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-%28-31%29%2B-sqrt%28%28-31%29%5E2-4%281%29%2884%29%29%29%2F%282%281%29%29

x+=+%2831%2B-sqrt%28961+-+336%29%29%2F%282%29

x+=+%2831%2B-sqrt%28625%29%29%2F%282%29

x+=+%2831%2B-++25%29%2F%282%29

x+=+%2831%2B25%29%2F%282%29 or x+=+%2831-25%29%2F%282%29

x+=+%2856%29%2F%282%29 or x+=+%286%29%2F%282%29

x+=+28 or x+=+3

If x = 28, then 27-2x = 27-2*28 = -29, which is not valid.
The side length must be positive.

If x = 3, then,
27-2x = 27-2*3 = 21
and
35-2x = 35-2*3 = 29
Both results are positive.
Therefore x = 3 is valid and practical.
A gap of 3 feet produces a rug that is 21 feet by 29 feet.

As a check, 21*29 = 609, and we have confirmed the answer is correct.

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Answer: The rug should be 21 feet by 29 feet.

Answer by josgarithmetic(39617) About Me  (Show Source):
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Problem is much like this one:
https://youtu.be/lM-Vkn9qMaI rectangular outside border of uniform width
https://youtu.be/lM-Vkn9qMaI

Answer by MathLover1(20849) About Me  (Show Source):
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Let x+be the width of the uniform strip floor around the rug.

Then the width of the rug is %2827+-+2x%29+and the length of the rug is %2835+-+2x%29.

%2827+-+2x%29%2835+-+2x%29+=+609
4x%5E2+-+124x+%2B+945+=609
4x%5E2+-+124x+%2B+945+-609=0
4x%5E2+-+124x+%2B+336=0
4%28x%5E2+-31x+%2B+84%29=0
4+%28x+-+28%29+%28x+-+3%29+=+0
solutions:
x+=+3+
or
x+=+28 (rejected as x must be less that the width of the room)
Hence,
the width of the rug is+27+-+2%283%29+=+21ft
and
the length of the rug is 35+-+2%283%29+=+29ft


Answer by Edwin McCravy(20054) About Me  (Show Source):
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For enrichment, here is a different approach.



Let the width of the strip be x. 
The area of the floor is 27x35 = 945, and the area of the rug is 609,
So the total area of the strip is 945-609 = 336

As you see from the drawing above, the strip is made up of two
vertical 27-x by x thin rectangles and vertical two horizontal
rectangles. So we have the equation:

2x%2827-x%29%22%22%2B%22%222x%2835-x%29%22%22=%22%22336

54x-2x%5E2%2B70x-2x%5E2%22%22=%22%22336

-4x%5E2%2B124x-336%22%22=%22%220

Divide through by -4

x%5E2-31x-84%22%22=%22%220

That's the same quadratic equation the other tutors got, so the
rest is the same as theirs.  I'll stop here.

Edwin

Answer by greenestamps(13198) About Me  (Show Source):
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You have received several responses showing different formal solutions, or showing links to where you can see a solution.

In a situation where finding the solution quickly is important and a formal algebraic solution is not required, there are much faster paths to the solution than formal algebra.

The difference between the dimensions of the room is 8 feet. So the solution will be two numbers whose difference is 8 and whose product is 609.

With no algebra at all, you can solve the problem simply with mental arithmetic to find those two numbers.

609 = 3(203) = 3(7)(29) = (21)(29)

ANSWERS: 21 feet and 29 feet

There is also a quick algebraic shortcut that can get you to the solution after determining that you need two numbers whose difference is 8 and whose product is 609.

Given the difference of 8 between the two numbers, let the two numbers be x+4 and x-4. Then, since the product is 609,

(x+4)(x-4) = 609
x^2-16 = 609
x^2 = 625
x = 25

And again the two dimensions are 25+4 = 29 feet and 25-4 = 21 feet.