Apropane tank is designed in the shape of a cylinder with two hemispherical ends. the cylinder's height is twice its diameter, and the tank's capacity is 1,000 gal. what is the radius of the tank, to the nearest tenth of a foot? (hint: 1 ft3 7.48 gal)

Let
x> the cylinder's height
y> the  cylinder's diameter

we know that
x=2*y
[volume of the tank]=[volume of the cylinder]+[volume of two hemispherical ends]

step 1
find the volume of the cylinder
[volume of the cylinder]=pi*r²*h
r=y/2
h=x> 2*y
so
[volume of the cylinder]=pi*(y/2)²*(2*y)> pi*y³/2

step 2
find the volume of two hemispherical ends
but
two hemispherical ends is equals to one sphere
so
[volume of the sphere]=(4/3)*pi*r³
r=y/2
[volume of the sphere]=(4/3)*pi*(y/2)³> (1/6)*pi*y³

step 3
find the radius of the tank

we know that
1 ft³> 7.48 gal
X ft³> 1000 gal
X=1000/7.48> X=133.69 ft³

[volume of the tank]=[volume of the cylinder]+[volume of two hemispherical ends]
so
133.69=[pi*y³/2]+[(1/6)*pi*y³]> 133.69=y³*[(pi/2)+(pi/6)]
133.69=y³*[(2/3)*pi]> y³=133.69/[(2/3)*pi]> y³=63.83
y=∛63.83> y=4 ft

therefore
the radius is y/2> r=4/2> r=2 ft

the answer is
r=2 ft

Distance is always positive