Respuesta :
Let x represent the width of the "uniform strip."
Write formulas: one for the length of the rug, one for the width.
If the room is 20 ft wide, the rug width would have to be 20-2x, and the run length would be 26-2x.
Rug area is 432 sq ft, and this equals (rug length)(rug width), or
432 = (26-2x)(20-2x) = 520 -40x - 52x + 4x^2 and this is 432.
Subtr. 432 from both sides: 520 - 432 - 92x + 4x^2 = 0. This is a quadratic equation that could be solved in various ways.
4x^2 - 92x + 88 = 0. Let's reduce this by div. all terms by 4:
x^2 -23x + 22 = 0
Easily factored! Note that -1x and -22x add up to -23x, and that (-1)(-12) = 22. Thus, the factors are (x-1)(x-22) = 0, so that x = 1 or x = 22.
Remembering that x represents the strip width, we eliminate x = 22 and keep x = 1.
The rug dimensions should be (26-2) by (20-2), or 24 by 18 feet.
Check: Does this area come out to 432 sq ft? (24)(18) = 432/ YES
Write formulas: one for the length of the rug, one for the width.
If the room is 20 ft wide, the rug width would have to be 20-2x, and the run length would be 26-2x.
Rug area is 432 sq ft, and this equals (rug length)(rug width), or
432 = (26-2x)(20-2x) = 520 -40x - 52x + 4x^2 and this is 432.
Subtr. 432 from both sides: 520 - 432 - 92x + 4x^2 = 0. This is a quadratic equation that could be solved in various ways.
4x^2 - 92x + 88 = 0. Let's reduce this by div. all terms by 4:
x^2 -23x + 22 = 0
Easily factored! Note that -1x and -22x add up to -23x, and that (-1)(-12) = 22. Thus, the factors are (x-1)(x-22) = 0, so that x = 1 or x = 22.
Remembering that x represents the strip width, we eliminate x = 22 and keep x = 1.
The rug dimensions should be (26-2) by (20-2), or 24 by 18 feet.
Check: Does this area come out to 432 sq ft? (24)(18) = 432/ YES
The dimensions of the rug should be 18 feet and 24 feet
The given parameters;
dimension of the room = 20 ft by 26 ft
maximum area of rug she can afford = 432 ft²
To find:
- the dimension of the rug
If they will be a uniform stripe of floor around the rug, then let the uniform excess length of the floor to removed from each dimension = y
[tex]Area = Length \times width\\\\432 = (20 - y ) \times (26-y)\\\\432 = 520 -20y -26y + y^2\\\\y^2 - 46y + 88 = 0\\\\\\solve \ the \ quadratic \ equation \ using \ formula \ method;\\\\a = 1, \ b = - 46, \ c = 88\\\\y = \frac{-b \ \ +/- \ \ \sqrt{b^2 -4ac} }{2a} \\\\y = \frac{-(-46) \ \ +/- \ \ \sqrt{(-46)^2 -4(1\times 88)} }{2(1)} \\\\y = 44 \ \ \ or \ \ \ 2[/tex]
The value of y cannot be greater than any of the original dimension, we will choose y = 2
The uniform dimension of the floor to be covered by the maximum area of rug she can afford = (20 - 2) and (26 - 2) = 18 and 24
Thus, the dimensions of the rug should be 18 feet and 24 feet
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