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February 1, 2015

February 1, 2015

Posted by **George** on Saturday, October 25, 2008 at 10:48pm.

owner wants 5-foot wide decks along either side and 10-foot wide decks at the two ends.

Find the dimensions of the smallest piece of property on which the pool can be built

satisfying these conditions.

Is the answer 16 by 18. Any help woulb be greatly appreciated.

- Calculus -
**drwls**, Sunday, October 26, 2008 at 9:23amWith 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

- Calculus -
**Reiny**, Sunday, October 26, 2008 at 9:28amYou 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

- Calculus -
**Jew**, Monday, January 25, 2010 at 9:54pm80 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.

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