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# The rectangle above is a diagram of the fencing surrounding an animal

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Math Expert
Joined: 02 Sep 2009
Posts: 47037
The rectangle above is a diagram of the fencing surrounding an animal [#permalink]

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22 Sep 2017, 00:36
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4
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Difficulty:

65% (hard)

Question Stats:

46% (01:44) correct 54% (01:36) wrong based on 35 sessions

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The rectangle above is a diagram of the fencing surrounding an animal pen 20 meters by 5 meters. If an additional 4 square meters is to be enclosed by moving just one side of the pen, while retaining its rectangular shape, what is the least possible number of meters of additional fencing needed?

(A) 2.0
(B) 1.6
(C) 1.0
(D) 0.8
(E) 0.4

Attachment:

2017-09-20_1023_002.png [ 2.6 KiB | Viewed 655 times ]

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Joined: 02 Jul 2017
Posts: 294
GMAT 1: 730 Q50 V38
Re: The rectangle above is a diagram of the fencing surrounding an animal [#permalink]

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22 Sep 2017, 02:45
Current Area = 20*5 = 100 sq meters
Additional area = 4 sq meter
Total area required 104 sq meters

Find : what is the least possible number of meters of additional fencing needed?

So if we move 5 meter fence :
so new width = new area/ length = 104/20 = 5.2
=> Therefore new measurement of rectangle = 20 * 5.2
Extra fencing needed = new width - old width *2 = 0.2*2 =0.4 meters ( Multiplying by 2 as width is increased at both ends)

And as this is the smallest number present in list we didn't tried moving length and then calculating new measurements

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Joined: 22 May 2016
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The rectangle above is a diagram of the fencing surrounding an animal [#permalink]

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22 Sep 2017, 16:36
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Bunuel wrote:

The rectangle above is a diagram of the fencing surrounding an animal pen 20 meters by 5 meters. If an additional 4 square meters is to be enclosed by moving just one side of the pen, while retaining its rectangular shape, what is the least possible number of meters of additional fencing needed?

(A) 2.0
(B) 1.6
(C) 1.0
(D) 0.8
(E) 0.4

Attachment:
2017-09-20_1023_002.png

Area increases from 100 to 104 $$m^2$$. One or both lengths must increase.

The "least possible number of meters of additional fencing needed" to obtain the additional area can be reframed: maximize the rectangle's area and minimize its perimeter.

The least change in perimeter to obtain the maximum change in area will come from adding length to the shorter sides.

That is, we need add the fencing to the shorter sides with length 5, because:
---The minimum perimeter that yields maximum area for a rectangle is a square.
---Extend that principle: in order for an increase in area to be obtained with the least addition to the perimeter ("least number of meters of fencing"), this rectangle needs to get closer to a square in shape.

If we extend the longer side of 20, the rectangle becomes even more oblong = less like a square

Hence, take the shorter side with length 5. Both sides must be extended. Although both short sides are extended, to calculate area, only one of the two short sides is used.

Let $$x$$ = additional meters of fencing needed

$$\frac{1}{2}x$$ gets added to each side with length 5

5 + $$\frac{1}{2}x$$ = New length of ONE short side, which gets multiplied by 20 to equal the new area

New area = $$(20)(5 +$$ $$\frac{1}{2}x$$)

$$104 = 100 + 10x$$

$$4 = 10x$$

$$\frac{4}{10} = 0.4 = x$$

P.S. By contrast, if you add $$\frac{1}{2} x$$ to the (long) sides with length 20, the minimum possible additional meters of fencing = 1.6
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The rectangle above is a diagram of the fencing surrounding an animal   [#permalink] 22 Sep 2017, 16:36
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