A spherical balloon is being inflated at the rate of 12 cubic feet per second. What is the radius of the balloon when its surface area is increasing at a rate of 8 feet square feet per second? Volume= (4/3)(pi)r^3 Area= 4(pi)r^2?

1 Answer
Alan P.
Apr 8, 2015

Let
#S# be the surface area (in square feet);
#V# be the volume of the sphere (in cubic feet);
#t# be the time (in seconds); and
#r# be the radius (in feet).

#(d S)/(dt) = 8# (given)
#rarr (dt)/(d S) = 1/8#
#(dV)/(dt) = 12# (given)
Therefore
#(dV)/(d S) = (d V)/(dt) * (dt)/(dS) = (12)/(8) = 3/2#

Using the Volume and Surface Area formulae (as given)
#(dV)/(dr) = 4 pi r^2#
and
#(dS)/(dr) = 8 pi r#
#rarr (dr)/(dS) = 1/(8 pi r)#
Therefore
#(dV)/(dS) = (dV)/(dr) * (dr)/(dS) = (4 pi r^2)/(8 pi r) = r/2#

Equating our two solutions for #(dV)/(dS)#
#r/2 = 3/2#

and the radius (#r#) is 3 feet.

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