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# M05-18

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Math Expert
Joined: 02 Sep 2009
Posts: 43853

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15 Sep 2014, 23:25
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Difficulty:

5% (low)

Question Stats:

78% (00:31) correct 22% (00:40) wrong based on 170 sessions

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To build a rectangular chicken pen, Mike has 40 meters of netting. If Mike wants to maximize the area of the pen, what will be the most favorable dimensions?

A. 12 x 8
B. 15 x 8
C. 10 x 10
D. 15 x 15
E. 15 x 5
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 43853

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15 Sep 2014, 23:25
Expert's post
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BOOKMARKED
Official Solution:

To build a rectangular chicken pen, Mike has 40 meters of netting. If Mike wants to maximize the area of the pen, what will be the most favorable dimensions?

A. 12 x 8
B. 15 x 8
C. 10 x 10
D. 15 x 15
E. 15 x 5

Approach 1:

Say the length and the width of the rectangle are $$x$$ and $$y$$.

Given: $$\text{Perimeter}=2x+2y=40$$. Reduce by 2: $$x+y=20$$. We have to maximize the area, so maximize the value of $$xy$$.

Useful property: for given sum of two numbers, their product is maximized when they are equal. Hence, the value of $$xy$$ will be maximized for $$x=y=10$$.

Approach 2:

This question can be easily solved if one knows the following property: a square has a larger area than any other quadrilateral with the same perimeter.

So, in order to maximize the area our rectangle must be a square: $$\text{Perimeter}=4x=40$$. Therefore $$x=10$$.

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Joined: 05 Aug 2015
Posts: 55

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12 Mar 2016, 23:12
Bunuel wrote:
Official Solution:

To build a rectangular chicken pen, Mike has 40 meters of netting. If Mike wants to maximize the area of the pen, what will be the most favorable dimensions?

A. 12 x 8
B. 15 x 8
C. 10 x 10
D. 15 x 15
E. 15 x 5

Approach 1:

Say the length and the width of the rectangle are $$x$$ and $$y$$.

Given: $$\text{Perimeter}=2x+2y=40$$. Reduce by 2: $$x+y=20$$. We have to maximize the area, so maximize the value of $$xy$$.

Useful property: for given sum of two numbers, their product is maximized when they are equal. Hence, the value of $$xy$$ will be maximized for $$x=y=10$$.

Approach 2:

This question can be easily solved if one knows the following property: a square has a larger area than any other quadrilateral with the same perimeter.

So, in order to maximize the area our rectangle must be a square: $$\text{Perimeter}=4x=40$$. Therefore $$x=10$$.

Isn't this problem flawed? The prompt specifically asks for a RECTANGULAR pen, so I excluded C and D b/c they are SQUARE pens??

Edit: OK, so Square is a special kind of Rectangle. Keeping this post as this is an important subtle point here for this problem.
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Math Expert
Joined: 02 Sep 2009
Posts: 43853

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13 Mar 2016, 07:17
happyface101 wrote:
Bunuel wrote:
Official Solution:

To build a rectangular chicken pen, Mike has 40 meters of netting. If Mike wants to maximize the area of the pen, what will be the most favorable dimensions?

A. 12 x 8
B. 15 x 8
C. 10 x 10
D. 15 x 15
E. 15 x 5

Approach 1:

Say the length and the width of the rectangle are $$x$$ and $$y$$.

Given: $$\text{Perimeter}=2x+2y=40$$. Reduce by 2: $$x+y=20$$. We have to maximize the area, so maximize the value of $$xy$$.

Useful property: for given sum of two numbers, their product is maximized when they are equal. Hence, the value of $$xy$$ will be maximized for $$x=y=10$$.

Approach 2:

This question can be easily solved if one knows the following property: a square has a larger area than any other quadrilateral with the same perimeter.

So, in order to maximize the area our rectangle must be a square: $$\text{Perimeter}=4x=40$$. Therefore $$x=10$$.

Isn't this problem flawed? The prompt specifically asks for a RECTANGULAR pen, so I excluded C and D b/c they are SQUARE pens??

Edit: OK, so Square is a special kind of Rectangle. Keeping this post as this is an important subtle point here for this problem.

Yes, every square is a rectangle but not vise-versa.
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Joined: 26 Dec 2016
Posts: 20

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28 Feb 2017, 05:59
Hi Bunuel,

it's may a stupid question, but how could you build a rectangle with the dimension 15x15 if you have only 40m netting ?

Thanks
VP
Joined: 22 May 2016
Posts: 1339

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25 Apr 2017, 04:58
BoomHH wrote:
Hi Bunuel,

it's may a stupid question, but how could you build a rectangle with the dimension 15x15 if you have only 40m netting ?

Thanks

IMO, not a dumb question, though you missed answer B's similar impossibility .

My VERY non-expert guess is that "15x15" is designed to hijack the person who, per Bunuel 's second explanation, correctly deploys "max area is square" but then, without tracking on actual numbers, incorrectly decides that if a square is good, a bigger square is better.
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M05-18   [#permalink] 25 Apr 2017, 04:58
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# M05-18

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