<PRE><B>Question 743450</B>
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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