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Re: math hw question

Posted: Tue Aug 11, 2009 10:54 pm
by Xeno
I thought I would have a go at this, although my method was similar to Avansc's I figured it might help out to post how I did it.

Ok, so firstly I made a diagram like so:

Image

Since the question tells you the pool's area is 1800m^2 then you know that
L*W = 1800

putting this in terms of a single variable, I chose L, gives
L = 1800 / W.

Now, because you are trying to find the area of the whole property (that is, the large bordering rectangle in my badly drawn diagram) you need another equation. Here I am using y for the area of the property:
y = ( W + 10 )( L + 20 )

Next I substituted in the earlier equation so that there are only 2 variables present.
y = ( W + 10 )(( 1800/W ) + 20 )

Expanding this gives
y = 20W + 18000/W + 2000

Now, to find the minimum point you can differentiate y then set dy/dW to 0 like so...

dy/dW = 20 -18000/(W^2)
dy/dW = 0
0 = 20 - 18000/(W^2)
18000/(W^2) = 20

Now you can solve this to find W^2 and hence W...
W^2 = 18000/20
W^2 = 900
W = +/- 30

since -30 is not a sensible answer for the width of a pool, take +30 instead, and then you can plug this value back in the original equations to find y.
W = 30
L * W = 1800
L = 60

y = ( W + 10 )( L + 20 )
y = ( 30 + 10 )( 60 + 20 )
y = 40*80
y = 3200.

Therefore the minimum area is 3200m^2

EDIT: I realised the question asked for the dimensions not the area, so 60m long by 30m wide would be the answer.