Question 945694
The town swimming pool is d feet deep.
 The width of the pool is 10 feet greater than 5 times the depth.
w = 5d+10
 The length of the pool is 25 feet greater than the width.
L = w + 25
Replace w with (5d+10)
L = (5d+10) + 25
L = 5d + 35
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A.Write and simplify an expression to represent the volume of the pool.
Volume = L*w*d
replace L and w
V = (5d+35)*(5d+10)* d
V = (25d^2 + 50d + 175d + 350)* d
V = (25d^2 + 225d + 350)*d
Factor out 25 and you have
V = 25d(d^2 + 9d + 14)
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B. If the pool holds 51,000 ft3 of water, what are the dimensions of the pool?
25d(d^2 + 9d + 14) = 51000
divide both sides by 25
d(d^2 + 9d + 14) = 2040
the equation
d^3 + 9d^2 + 14d - 2040 = 0
Using the Ti83, I got d = 10 ft is the depth
then
5(10) + 10 = 60 ft is the width
and
5(10) + 35 = 85 ft is length
:
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Check this, find the vol using these values; 85 * 60 * 10 = 51000