Answer to Question #272446 in Calculus for Paul

Question

The dimensions of a box are b, b+1, b+4. Find how fast the volume increases as b increases

Answer 1

volume is

v = b(b + 1)(b + 4) = b³ + 5b² + 4b

differentiating both sides w.r.t. t,

we get dv/dt = dv/db * db/dt

=3b².db/dt + 10b.db/dt + 4db/dt

dv/dt = (3b² + 10b + 4).db/dt

so, the rate is dv/dt = (3b² + 10b +4 )*db/dt

OR,

directly calculate dv/db which is same,

v = b(b + 1)(b + 4) = b³ + 5b² + 4b

dv/db = (3b² + 10b +4 )