Math Question

Discussion in 'Teacher Time Out' started by ku_alum, Mar 21, 2009.

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Mar 21, 2009

My husband is building me a raised bed vegetable garden (google square foot gardening if you're intersted). Anyway ... if he uses 8 foot boards and builds it 4 feet by 4 feet we have 16 square feet. If he uses the SAME materials, but builds it 3 feet by 5 feet, we have 15 square feet.

Am I just having a blond moment? Why does it work that way, why do we lose a foot even though we are using the same materials??

2. 3. AliceaccMultitudinous

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Mar 21, 2009

Because 2 shapes can have the same perimeter but different areas.

With your 16 linear feet of perimeter, some options include:
1x7, area 7
2x6, area 12
3x5, area 15
4x4, area 16.

The square will always maximize the area.

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Mar 21, 2009

I'm not trying to be difficult ... but, why?

5. Catcherman22Companion

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Mar 21, 2009

lets assume each foot holds one plant.. so a 3 X 5 hold 3 plants one direction and 5 plants another direction..

start with the 4 by 4. 4 plans long and 4 plants wide.

x x x x
x x x x
x x x x
x x x x

so to go to a 3 x 5 you need to remove the 4th column (so you have 3 columns).

x x x x
x x x x
x x x x
x x x x

To get 5 rows, you need one of each of the removed column and put it on each remaining column.

x x x
x x x
x x x
x x x
x x x x

This leaves the one remaining plant out in the open.. and is hence lost.

That help?

6. MalcolmEnthusiast

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Mar 21, 2009

A=lw

4'x4' = 16 square feet

3'x5' = 15 square feet

A square will always maximize the area for a given perimeter in a rectangle.

The more different the side lengths are, the smaller will be the area for the same perimeter.

2'x6' = 12 square feet

1'x7' = 7 square feet

7. HMMCohort

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Mar 25, 2009

if you have a constraint for the perimeter 2W+2L=C

and you want to maximize the area A=L*W....you can use a little calculus

from 2W+2L=C we can get

W=(C-2L)/2

plug into A=L*W

so

A=L*(C-2L)/2=(1/2)(C*L)-L^2

findthe derivative of A with respect to L

A'=C/2-2L

and set equal to 0 and solve for L

C/2=2L
L=C/4

then W=(C-2L)/2=(C-2C/4)/2=(C/2)/2=C/4

so L=W

and this is a max since

A''=-2<0 for any value of L (2nd derivative test)

you could have used Lagrange Multiplies if you wanted to but that would have been a little overkill

8. TeacherGroupieModerator

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Mar 25, 2009

HMM, I'm afraid I could feel ku_alum's eyes go glassy all the way over on the Left Coast...

Neatly explained, Catcherman.

9. HMMCohort

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Just thought I'd share 