Calculating W, L, r, and x for tile, before pool is built

Y

yellowmonster572

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Sep 12, 2007
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Mrs. I. M. Mean has hired you [a tile contractor] to tile a walking area around a small wading pool about to be built at the rear of her house.

The custom tile she has purchased and will use is very expensive, costing $22.00 per square inch. Since the tile is custom, there is no additonal tile available. She has purchased 300 square feet of the tile and requires you to use ALL the tile in the construction of the walkway.

Before the wading pool is built, she asks you to calculate:

-the values of W, L, and r.
-the value of "x" [the walkway width].

I have a sketch of this problem showing that the pool has a semi circular end, causing 'r' to be in the problem.

The length of the area the pool/walkway will be in is 39' 4" and the width is 21' 2".

So far I have
(39.33)(21.17)=(lw+[pi r^2/over2])+300
2x+w=21.17
2x+l+r=39.33

Where do I go from here?
 
Re: Word Problem

yellowmonster572 said:
Mrs. I. M. Mean has hired you [a tile contractor] to tile a walking area around a small wading pool about to be built at the rear of her house.

The custom tile she has purchased and will use is very expensive, costing $22.00 per square inch. Since the tile is custom, there is no additonal tile available. She has purchased 300 square feet of the tile and requires you to use ALL the tile in the construction of the walkway.

Before the wading pool is built, she asks you to calculate:
-the values of W, L, and r.
-the value of "x" [the walkway width].


I have a sketch of this problem showing that the pool has a semi circular end, causing 'r' to be in the problem.


The length of the area the pool/walkway will be in is 39' 4" and the width is 21' 2".

Why is length(L) and width (W) given - since she asked you to calculate the values of W,L and r?

So far I have
(39.33)(21.17)=(lw+[pi r^2/over2])+300
2x+w=21.17
2x+l+r=39.33


Where do I go from here?
Click to expand...

The pool has One semicircular end - or two semicircular ends?
 
Is this what the pool looks like?. Please excuse it's rough appearance. I used "paint". :roll: :lol:

Please post the problem statement exactly as it appears.


pooltileoy1.gif
 
Yikes YM, that's quite the problem! I'll use P for pi.

First, use fractions: 39'4" = 118/3, 21'2" = 127/6
So total area = 118/3 * 127/6 = 7493/9

A bit like you have:
7493/9 - 300 = WL + Pr^2 / 2
But W = 2r (right?); so:
7493/9 - 300 = 2Lr + Pr^2 / 2
Simplifies to:
9Pr^2 + 36Lr - 9586 = 0 [1]

OK; total width = 2x + 2r; so:
2x + 2r = 127/6
Simplifies to:
x = (127 - 12r) / 12 [2]

Total length = L + 2x + r; so:
L + 2x + r = 118/3
Simplifies to:
x = (118 - 3L - 3r) / 6 [3]

Using [2] and [3]:
(127 - 12r) / 12 = (118 - 3L - 3r) / 6
Simplifies to:
L = (6r + 109) / 6 [4]

Substitute [4] in [1]:
9Pr^2 + 36r((6r + 109) / 6) - 9586 = 0
9Pr^2 + 36r^2 + 654r - 9586 = 0
(9P + 36)r^2 + 654r - 9586 = 0

Let k = 9P + 36 and use the quadratic:
r = [-654 + sqrt(654^2 - 4(k)(-9586))] / (2k)

I'll give you the fun of finishing that off; you should get r = ~8.1421
Substitute back to get the other unknowns.

Yer lucky I'm in a good mood: my grandson Tyler (15) made the
Ottawa Wolverines competitive basketball team tonight: and in age 16-17 division!
 
Denis said:
Yikes YM, that's quite the problem! I'll use P for pi.

First, use fractions.....

Yer lucky I'm in a good mood: my grandson Tyler (15) made the
Ottawa Wolverines competitive basketball team tonight: and in age 16-17 division!
Click to expand...
Thank you so much!! And tell Tyler that I send him congrats, lol.
 
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