Frame Problem - Need help turning given information into a quadratic equation.

codyw1996

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Mar 6, 2017
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Okay so this question is a frame problem meant to be solved by turning the problem into a quadratic equation.

I'm having trouble extracting enough information from the problem. Or maybe I just don't know how to use the given information.

Here's the word problem:

A landscape architect is designing a rectangular flowerbed with 28 plants that are placed 1 meter apart.
He needs an inner rectangular space in the center for plants that must be 1 meter from the border of the bed
and that require 24 square meters for planting. What should the overall dimensions of the flowerbed be?

Ok, so here's what I think i've gathered from this problem.

All plants are placed 1 meter apart, but there are 2 different kinds of plants.
One kind forms the inner rectangle that has an area of 24 square meters, and they
must be 1 meter from the border, making the border width (frame width) 1.
The rest of the plants surround the inner rectangle forming a larger rectangle - total
area 28 square meters.

So, all I can confirm is that:

x=1 ;frame width
lw = 24 ;area of inner rectangle
(l+2x)(w+2x)=28 ;entire area, x = 1

rotated3.jpg

I don't know how to turn this info into a quadratic that will give me the dimensions of the flowerbed.
The dimensions being: 6m x 8m

Obviously 6 x 8 = 48, meaning that the entire area must be 48 square meters, not 28.
And it maks sense, because (6-2(1))(8-2(1)) = 4 x 6 = 24. It works.

Possibly a typo, maybe it was supposed to be 48 total flowers instead of 28. If that's the case, I could have solved it
just by looking at it and thinking about the factors of 24.

But it's still possible that i'm completely misinterpreting the question.

Thanks.
 
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Cody

The problem as stated is not very clear. If that is the way the problem was posed, I give low marks to the writer.

However, if we make a few assumptions that APPEAR WARRANTED by the language of the problem AS YOU HAVE GIVEN it, it is soluble. But perhaps you did not give the language of the problem completely and accurately. Furthermore, you have definitely made some erroneous assumptions.

We have 28 plants. As you have specified it, the problem does NOT require that the area of the flower bed be 28 square meters. You apparently added that twist on your own. So what is required by what I am reading as the problem is that the inner rectangle be 24 square meters, but we are to find the dimensions of the whole flower bed. The presumption is that they want those dimensions to minimize the outer rectangle' s area because otherwise there are an infinite number of valid answers. Moreover, there is no guarantee that the minimum will be the same for every layout of the inner rectangle. Finally, nothing is said explicitly about two types of plant, but I agree that such a reading makes sense.

Type I plants: must be planted 1 meter away from any other plant and from the outer border.

Type II plants can be planted anywhere.

There are 24 plants of type I and 4 plants of type II though the problem does not say so explicitly. (As I said, whoever wrote this problem should be shot.)

Your big fallacy here is that the outer border must be 1 meter from the inner border. Not so.

Let's say we divide the inner rectangle of 24 square meters into x rows and y columns of squares each with sides of 1 meter. How are they laid out? Well the problem apparently does not tell us (or you did not tell us).

Note \(\displaystyle xy = 24\ and\ x,\ y \in \mathbb Z^+\).

That means we could have eight different possibilities, namely 24, 1 or 12, 2 or 8, 3 or 6, 4 or 4, 6 or 3, 8, or 2, 12, or 1, 24.

So we plant 1 of type I in the center of each square. That makes them all 1 meter away from each other and gives them each 1 square meter of space. So far so good. And no matter where we set up the outer border, the squares adjacent to it are ALREADY 0.5 meters away from the outer border. So the outer border needs to be only 0.5 more meters away from the inner border to get every plant of type I at least 1 meter from the outer border.

Where do we plant the 4 type II plants? In the four corners of the outer rectangle. Will that work? Yes the distance from the center of any square located at the corner of the inner rectangle to the nearest corner of the outer rectangle will be

\(\displaystyle \sqrt{(0.5 + 0.5)^2 + (0.5 + 0.5)^2} = \sqrt{1^2 + 1^2} = \sqrt{2} > 1.4 > 1.\)

Now what are the dimensions of the outer rectangle: obviously

\(\displaystyle (x + 2 * 0.5) = (x + 1)\ and\ (y + 2 * 0.5) = (y + 1).\)

The resulting area is

\(\displaystyle (x + 1)(y + 1) = xy + x + y + 1 = 24 + x + y + 1 = 25 + x + y.\)

The possible sums of x and y are 25, 14, 11, and 10. So we minimize the area by setting x = 6 and y = 4 or x = 4 and y = 6.

That gives dimensions for the outer rectangle of 7 meters by 5 meters and a minimum area of 35 square meters.

EDIT: I am not sure I see where solving a quadratic equation is involved.
 
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Denis, that is a reasonable guess as to what the problem was. We cannot be sure whether the actual problem was atrociously worded or whether the student's paraphrase was confused.
 
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