SOLUTION: A customer wants an aquarium that is exactly twice as long as it is wide and holds 30 gallons of water . Suggest at least two sets of dimensions to the nearest tenth inch that woul

Question 743450: A customer wants an aquarium that is exactly twice as long as it is wide and holds 30 gallons of water . Suggest at least two sets of dimensions to the nearest tenth inch that would work for this combination. Here is some more info...
Common aquarium holds 10 gallons. I know volume of a right rectangular tank is found by multiplying the length of the tank times the width times the height, or V=lwh. I know 231 cubic inches are equal to 1 gallon.
I am TERRIBLE at word problems especially this one. Can you please help me?

Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
The problem states:
A customer wants an aquarium that is exactly twice as long as it is wide and holds 30 gallons of water . Suggest at least two sets of dimensions to the nearest tenth inch that would work for this combination.
we are given:
length(l) = 2*width(w)
we can write the formula for volume(v), note that volume is in cubic feet
v = 2w * w * depth(d)
now we know that the volume * 7.48 (number of gallons in a cubic foot) = 30 gallons
so the volume in cubic feet = 30 / 7.48 = 4
now we have
4 = 2w^2*d
convert the 4 cubic feet to cubic inches
4 * 1728 = 6912 cubic inches
and
6912 = 2w^2*d
3456 = w^2*d
w = (sqrt (3456/d) = 58.8 / sqrt (d)
if d = 16 inches, then w = 14.7 and the length is 29.4 inches
if d = 36 inches, then w = 9.8 and the length is 19.6 inches

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