SOLUTION: A club swimming pool is 17 ft wide and 33 ft long. The club members want an exposed aggregate border in a strip of uniform width around the pool. They have enough material for 600

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Question 1101091: A club swimming pool is 17 ft wide and 33 ft long. The club members want an exposed aggregate border in a strip of uniform width around the pool. They have enough material for 600 ft squared ft 2. How wide can the strip​ be?
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let w be a uniform strip width.


Then the outside (exterior) dimensions of the larger rectangle are 17+2w and 33+2w.


The area of the larger rectangle is (17+2w)*(33+2w).


The equation for the area of the border strip around the pool is

(17+2w)*(33+2w) - 17*33 = 600,   or

34w + 66w + 4w^2 = 600,

4w^2 + 100w - 600 = 0  ====>

w^2 + 25w - 150 = 0,

Factor left side:

(w+30)*(w-5) = 0

The roots are w= -30  and  w= 5.


Only positive root makes sense:  w = 5 ft.


Answer.  The strip should be 5 ft wide.


To see more similar solved problems, look into the lesson
    - Problems on the area and the dimensions of a rectangle surrounded by a strip
in this site.