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I have a long rectangular metal sheet of 12' wide and I need

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I have a long rectangular metal sheet of 12' wide and I need [#permalink] New post 11 Jul 2003, 04:59
I have a long rectangular metal sheet of 12' wide and I need to construct a rain gutter by turning up two sides so that they are perpendicular to the sheet. Help me, all wonderful mathematicians out there, to find out how many inches should be turned up to give my rain gutter to hold its greatest capacity ?

:help2
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 [#permalink] New post 11 Jul 2003, 05:15
try out the combinations

2*5 =10
4*4 =16
5*3.5 =17.5
6*3 = 18
7*2.5 = 17.5
8*2 =16


maximum is at 18, so it should be turned up 3"
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 [#permalink] New post 11 Jul 2003, 05:31
No problem, Sir

You have to bend your metal sheet in some way, to have a П-shaped object. Since, the lenght is not important, concentrate on a П-section.
A section of maximal area will give you maximal volume.

So, how to bend a 12' line in 3-side, П-object to cover the maximal area?
Imagine a rectangular with sides: X, X, 12–2X, 12–2X. Let X be a vertical side, and 12–2X be a horizontal one.

X+X+12–2X=given 12 and one side is virtual

An area of the section is X*(12–2X)= 12X–2X^2
A derivative is 12–4X=0, where X=3

so your gutter should have two 3' vertical sides and one 6' horizontal.
Such section gives you 18 square inch area

All others are smaller (10+1+1) gives 10
(8+2+2) gives 16
(4+4+4) gives 16

With the same perimetr a circle has the smallest area, but a square -- the greatest.
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 [#permalink] New post 11 Jul 2003, 05:45
stolyar wrote:
No problem, Sir

You have to bend your metal sheet in some way, to have a П-shaped object. Since, the lenght is not important, concentrate on a П-section.
A section of maximal area will give you maximal volume.

So, how to bend a 12' line in 3-side, П-object to cover the maximal area?
Imagine a rectangular with sides: X, X, 12тАУ2X, 12тАУ2X. Let X be a vertical side, and 12тАУ2X be a horizontal one.

X+X+12тАУ2X=given 12 and one side is virtual

An area of the section is X*(12тАУ2X)= 12XтАУ2X^2
A derivative is 12тАУ4X=0, where X=3

so your gutter should have two 3' vertical sides and one 6' horizontal.
Such section gives you 18 square inch area

All others are smaller (10+1+1) gives 10
(8+2+2) gives 16
(4+4+4) gives 16

With the same perimetr a circle has the smallest area, but a square -- the greatest.


Gotcha, So I all I need to do is find the maximum value for the qudratic function representing the area of П-section.
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 [#permalink] New post 11 Jul 2003, 05:48
brstorewala wrote:
try out the combinations

2*5 =10
4*4 =16
5*3.5 =17.5
6*3 = 18
7*2.5 = 17.5
8*2 =16


maximum is at 18, so it should be turned up 3"


Excellent idea to minimize the time in actual test !! Agreed
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 [#permalink] New post 11 Jul 2003, 10:09
stolyar wrote:
No problem, Sir

You have to bend your metal sheet in some way, to have a П-shaped object. Since, the lenght is not important, concentrate on a П-section.
A section of maximal area will give you maximal volume.

So, how to bend a 12' line in 3-side, П-object to cover the maximal area?
Imagine a rectangular with sides: X, X, 12тАУ2X, 12тАУ2X. Let X be a vertical side, and 12тАУ2X be a horizontal one.

X+X+12тАУ2X=given 12 and one side is virtual

An area of the section is X*(12тАУ2X)= 12XтАУ2X^2
A derivative is 12тАУ4X=0, where X=3

so your gutter should have two 3' vertical sides and one 6' horizontal.
Such section gives you 18 square inch area

All others are smaller (10+1+1) gives 10
(8+2+2) gives 16
(4+4+4) gives 16

With the same perimetr a circle has the smallest area, but a square -- the greatest.


An area of the section is X*(12тАУ2X)= 12XтАУ2X^2
A derivative is 12тАУ4X=0, where X=3

This does not look correct!
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 [#permalink] New post 11 Jul 2003, 11:19
kpadma wrote:
stolyar wrote:
No problem, Sir

You have to bend your metal sheet in some way, to have a П-shaped object. Since, the lenght is not important, concentrate on a П-section.
A section of maximal area will give you maximal volume.

So, how to bend a 12' line in 3-side, П-object to cover the maximal area?
Imagine a rectangular with sides: X, X, 12тАУ2X, 12тАУ2X. Let X be a vertical side, and 12тАУ2X be a horizontal one.

X+X+12тАУ2X=given 12 and one side is virtual

An area of the section is X*(12тАУ2X)= 12XтАУ2X^2
A derivative is 12тАУ4X=0, where X=3

so your gutter should have two 3' vertical sides and one 6' horizontal.
Such section gives you 18 square inch area

All others are smaller (10+1+1) gives 10
(8+2+2) gives 16
(4+4+4) gives 16

With the same perimetr a circle has the smallest area, but a square -- the greatest.


An area of the section is X*(12тАУ2X)= 12XтАУ2X^2
A derivative is 12тАУ4X=0, where X=3

This does not look correct!


Brush up your calculus basics !! ( GMAT does not test u in calculus)

d (12x-x^2)
---- = 12-4x, right ?
dx

maximum value is 3 as you see from the solution
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 [#permalink] New post 11 Jul 2003, 13:18
Thanks!
:oops:
  [#permalink] 11 Jul 2003, 13:18
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I have a long rectangular metal sheet of 12' wide and I need

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