Posted by George on Saturday, October 25, 2008 at 10:48pm.
With lot dimensions of 16 x 18, you don't even have enough room for the pool. How would you fit in an 1800 square feet of pool?
You need to set this up as a calculus problem and look for the minimum of the Lot Area function. Let x be the shorter width of the pool rectangle. The other pool lendth dimension is 1800/x. The 10 foot deck widths should be on the narrower "x" sides of the pool, to minimize total area.
Total land area needed =
A(x) = 1800 + 2*10(x+10)+ 2*5(1800/x)
= 1800 + 20x + 200 + 18000/x
= 2000 + 20 x + 18000/x
Set the derivative equal to zero and solve for x.
20 - 18000/x^2 = 0
x = sqrt (900) = 30
pool length dimension = 1800/x = 60
Lot dimensions: 80 x 40
You would know your answer is illogical, since 16x18 is only 288 square feet!!
Let the pool be x feet by y feet.
then the property must be x+20 feet by y+10 feet.
and the area = (x+20)(y+10)
but xy= 1800 ---> y = 1800/x
then
Area =(x+20)(1800/x+10)
= 1800 + 10x + 36000/x + 200
Area' = 10 - 36000/x^2
= 0 for a max/min of Area
solve this to get x=60
so the pool is 60 by 30
and the property must be 80 by 40
80 x 40
I can't figure it out on my own it's way too confusing. It makes me forget my basic geometry skills because it violates the properties for finding the area of a rectangle in so many ways.