SOLUTION: A club swimming pool is 24 ft wide and 40 ft long. The club members want an exposed aggregate border in a strip of uniform width around the pool. They have enough material

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Question 1115712: A club swimming pool is 24

ft wide and 40

ft long. The club members want an exposed aggregate border in a strip of uniform width around the pool. They have enough material for 576

ft squared
.
How wide can the strip​ be?
Found 3 solutions by stanbon, ikleyn, Shin123:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A club swimming pool is 24ft wide and 40ft long.
The club members want an exposed aggregate border in a strip of uniform width around the pool. They have enough material for 576ft squared.
How wide can the strip​ be?
--------------------------------
Length = 2(24+40) = 2(64) = 128
Width = x
------
Equation::
128x = 576
Width = x = 576/128 = 4.5 ft
-----------------------------
Cheers,
Stan H.
----------

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.

         * * * This absolutely standard problem on quadratic equation application was solved in this forum at least 100 times. * * * 


The larger rectangle dimensions are (24+2*x) feet and (40+2*x) feet.

The larger rectangle area is the product of its dimensions  (24+2*x)*(40+2*x) square feet.

The pool area is 24*40 square feet.


The area of the uniform border around the pool is the difference of the two areas  (24+2*x)*(40+2*x) - 24*40.


It is exactly  576 square feet, which gives you your governing equation


    (24+2x)*(40+2x) - 24*40 = 576


Simplify and solve it for x:


    2x*40 + 2x*24 + 4x^2 = 576


    4x^2 + 128x - 576 = 0


    x^2 + 32x - 144 = 0


    (x-4)*(x+36) = 0


The only positive root  x= 4 ft  is the solution.   ANSWER


CHECK.  (24+2*4)*(40+2*4) - 24*40 = 576.    ! Correct !

Solved.

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Answer by Shin123(626) About Me  (Show Source):
You can put this solution on YOUR website!
The length could be 2(24+40)=2(64)=128
width=x
128x=576/128=4.5 ft.